To determine the line of best fit (linear regression) for the given data points, we need to calculate the slope (m) and the y-intercept (b) of the line \( y = mx + b \).
Given the data points:
| x | y | |---|-----| | 1 | 5 | | 2 | 9 | | 3 | 10 | | 4 | 10.5| | 5 | 11 |
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Calculate the averages:
- \( \bar{x} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3 \)
- \( \bar{y} = \frac{5 + 9 + 10 + 10.5 + 11}{5} = \frac{45.5}{5} = 9.1 \)
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Calculate the slope (m): \[ m = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2} \]
- \(\sum (x_i - \bar{x})(y_i - \bar{y})\):
- For \(x=1, y=5\): \((1-3)(5-9.1) = (-2)(-4.1) = 8.2\)
- For \(x=2, y=9\): \((2-3)(9-9.1) = (-1)(-0.1) = 0.1\)
- For \(x=3, y=10\): \((3-3)(10-9.1) = (0)(0.9) = 0\)
- For \(x=4, y=10.5\): \((4-3)(10.5-9.1) = (1)(1.4) = 1.4\)
- For \(x=5, y=11\): \((5-3)(11-9.1) = (2)(1.9) = 3.8\)
- Sum: \(8.2 + 0.1 + 0 + 1.4 + 3.8 = 13.5\)
- \(\sum (x_i - \bar{x})^2\):
- For \(x=1\): \((-2)^2 = 4\)
- For \(x=2\): \((-1)^2 = 1\)
- For \(x=3\): \(0^2 = 0\)
- For \(x=4\): \(1^2 = 1\)
- For \(x=5\): \(2^2 = 4\)
- Sum: \(4 + 1 + 0 + 1 + 4 = 10\)
Calculating \(m\): \[ m = \frac{13.5}{10} = 1.35 \]
- \(\sum (x_i - \bar{x})(y_i - \bar{y})\):
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Calculate the y-intercept (b): \[ b = \bar{y} - m\bar{x} = 9.1 - 1.35 \cdot 3 \] \[ b = 9.1 - 4.05 = 5.05 \]
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Equation of the line: So the line of best fit is: \[ y = 1.35x + 5.05 \]
After rounding, the answer is:
y = 1.35x + 5.05
Thus, the correct response is:
- y=1.35x+5.05