The given linear regression equation is $y = 0.5x$, and the table has an observed value $y = 3$ for $x = 5$.

Let's break it down step by step:

1. Predicted Value: Using the equation $y = 0.5x$, substitute $x = 5$: $y_{\text{predicted}} = 0.5 \times 5 = 2.5$ So, the predicted value is 2.5.

2. Residual Value: The residual is the difference between the observed value and the predicted value: $\text{Residual} = y_{\text{observed}} - y_{\text{predicted}} = 3 - 2.5 = 0.5$

Completed Table:

Would you like more details or further clarification on any part of this solution?

Relative Questions:

1. What is the role of residuals in linear regression?
2. How does the slope affect the predicted values in a linear regression equation?
3. How would a larger residual value affect the accuracy of the model?
4. Can the residual value be negative, and what does it indicate?
5. How would the prediction change if the slope of the equation were steeper (e.g., $y = 1x$)?

Tip:

Residuals close to zero indicate that the regression model is a good fit for the data.