To find the least-squares regression line $\hat{y} = mx + b$, where $x$ is the number of absences and $y$ is the final grade, we need to calculate the slope $m$ and the intercept $b$ using the formulas:

$m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2}$ $b = \frac{\sum y - m(\sum x)}{n}$

where:

• $n$ is the number of data points,
• $\sum xy$ is the sum of the products of each pair of $x$ and $y$ values,
• $\sum x$ is the sum of the $x$-values,
• $\sum y$ is the sum of the $y$-values,
• $\sum x^2$ is the sum of the squares of the $x$-values.

Let's compute these values for the given data.

Data:

• $x$ (Number of absences): $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$
• $y$ (Final grade): $87.8, 84.9, 81.9, 79.3, 76.3, 71.9, 62.0, 66.7, 63.8, 60.8$

I'll now compute the necessary sums to find the least-squares regression line.The least-squares regression line is given by:

$\hat{y} = -3.17x + 87.80$

where:

• $x$ is the number of absences,
• $\hat{y}$ is the predicted final grade.

This equation indicates that for each additional absence, the final grade decreases by approximately 3.17 points.

Do you have any questions or would you like more details? Here are 5 related questions:

1. How can we use this regression equation to predict the final grade for 5 absences?
2. What is the interpretation of the slope in the context of this problem?
3. How would the regression line change if the number of absences had a smaller effect on the grades?
4. What is the predicted grade for a student with 0 absences?
5. Can we compute the residual for a student with 4 absences using this model?

Tip: In regression analysis, always check the residuals to assess how well the model fits the data!