Find the equation of the least-squares regression line for the given data. Round values to three decimal places. -5 -3 11 -6 X y OA.y=0.206x-2.097 OB. y=2.097x-0.206 C. y = -2.097x+ 0.206 D. y=-0.206x + 2.097 4 1 8 -3 -1 -2 -2 1 05 2 -5 3 6 -41 7

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## Finding the Equation of the Least-Squares Regression Line

**Given Data:**

- \( x \) values: -5, -3, 4, 1, -1, -2, 0, 2, 3, -4
- \( y \) values: 11, -6, 8, -3, -2, 1, 5, -5, 6, 7

**Task:** Find the equation of the least-squares regression line for the given data. Round values to three decimal places.

**Options:**

A. \( \hat{y} = 0.206x - 2.097 \)

B. \( \hat{y} = 2.097x - 0.206 \)

C. \( \hat{y} = -2.097x + 0.206 \)

D. \( \hat{y} = -0.206x + 2.097 \)

**Correct Answer:**
- The selected answer is option B: \( \hat{y} = 2.097x - 0.206 \)

### Explanation:
The least-squares regression line is used to predict the relationship between the independent variable \( x \) and the dependent variable \( y \). It minimizes the sum of the squares of the residuals (the differences between the observed and predicted \( y \) values). The equation is in the form \( \hat{y} = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In this scenario, the slope is approximately 2.097, and the y-intercept is approximately -0.206.
Transcribed Image Text:## Finding the Equation of the Least-Squares Regression Line **Given Data:** - \( x \) values: -5, -3, 4, 1, -1, -2, 0, 2, 3, -4 - \( y \) values: 11, -6, 8, -3, -2, 1, 5, -5, 6, 7 **Task:** Find the equation of the least-squares regression line for the given data. Round values to three decimal places. **Options:** A. \( \hat{y} = 0.206x - 2.097 \) B. \( \hat{y} = 2.097x - 0.206 \) C. \( \hat{y} = -2.097x + 0.206 \) D. \( \hat{y} = -0.206x + 2.097 \) **Correct Answer:** - The selected answer is option B: \( \hat{y} = 2.097x - 0.206 \) ### Explanation: The least-squares regression line is used to predict the relationship between the independent variable \( x \) and the dependent variable \( y \). It minimizes the sum of the squares of the residuals (the differences between the observed and predicted \( y \) values). The equation is in the form \( \hat{y} = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. In this scenario, the slope is approximately 2.097, and the y-intercept is approximately -0.206.
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