To calculate the correlation coefficient for the given data, we can use the Pearson correlation coefficient formula:
\[ r = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n \Sigma x^2 - (\Sigma x)^2][n \Sigma y^2 - (\Sigma y)^2]}} \]
where:
- \( n \) = number of data points
- \( \Sigma xy \) = sum of the product of paired scores
- \( \Sigma x \) = sum of x scores
- \( \Sigma y \) = sum of y scores
- \( \Sigma x^2 \) = sum of squared x scores
- \( \Sigma y^2 \) = sum of squared y scores
Let's denote the data points:
| x | y | |---|----| | 1 | 16 | | 4 | 20 | | 8 | 15 | | 6 | 22 | | 2 | 9 |
Now we can calculate the necessary sums.
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Calculate \( n \):
- \( n = 5 \)
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Calculate sums:
- \( \Sigma x = 1 + 4 + 8 + 6 + 2 = 21 \)
- \( \Sigma y = 16 + 20 + 15 + 22 + 9 = 82 \)
- \( \Sigma xy = (1 \cdot 16) + (4 \cdot 20) + (8 \cdot 15) + (6 \cdot 22) + (2 \cdot 9) = 16 + 80 + 120 + 132 + 18 = 366 \)
- \( \Sigma x^2 = 1^2 + 4^2 + 8^2 + 6^2 + 2^2 = 1 + 16 + 64 + 36 + 4 = 121 \)
- \( \Sigma y^2 = 16^2 + 20^2 + 15^2 + 22^2 + 9^2 = 256 + 400 + 225 + 484 + 81 = 1446 \)
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Substituting into the formula:
\[ r = \frac{5(366) - (21)(82)}{\sqrt{[5(121) - (21)^2][5(1446) - (82)^2]}} \]
Calculating each component step-by-step:
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Numerator: \[ 5(366) - (21)(82) = 1830 - 1722 = 108 \]
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Denominator:
- First part: \[ 5(121) - (21)^2 = 605 - 441 = 164 \]
- Second part: \[ 5(1446) - (82)^2 = 7230 - 6724 = 506 \]
- Now combine for the denominator: \[ \sqrt{(164)(506)} = \sqrt{82984} \approx 288.18 \]
Now calculating \( r \):
\[ r \approx \frac{108}{288.18} \approx 0.375 \]
Rounding to two decimal places gives \( 0.37 \).
Thus, the correlation coefficient for the data is approximately 0.37.
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