To solve this, we need to compute the squared errors for the regression line $Y = 2 + 3X$ using the given data points.

Step 1: Compute the predicted $Y$-values using the regression line formula $Y = 2 + 3X$.

• For $X = 0$: $Y_{\text{pred}} = 2 + 3(0) = 2$
• For $X = 3$: $Y_{\text{pred}} = 2 + 3(3) = 11$
• For $X = 7$: $Y_{\text{pred}} = 2 + 3(7) = 23$
• For $X = 10$: $Y_{\text{pred}} = 2 + 3(10) = 32$

Step 2: Compute the actual errors by subtracting the predicted $Y$-values from the given $Y$-values:

• For $X = 0$: $\text{Error} = 5 - 2 = 3$
• For $X = 3$: $\text{Error} = 5 - 11 = -6$
• For $X = 7$: $\text{Error} = 27 - 23 = 4$
• For $X = 10$: $\text{Error} = 31 - 32 = -1$

Step 3: Square each error:

• For $X = 0$: $(3)^2 = 9$
• For $X = 3$: $(-6)^2 = 36$
• For $X = 7$: $(4)^2 = 16$
• For $X = 10$: $(-1)^2 = 1$

Final Answer:

The squared errors are: $9, 36, 16, 1$.

Would you like more details or explanations? Here are 5 related questions to deepen your understanding:

1. How do you compute the residuals in linear regression?
2. What is the role of the sum of squared errors in determining the goodness of fit?
3. How can you adjust a regression model to reduce squared errors?
4. What is the difference between mean squared error (MSE) and sum of squared errors (SSE)?
5. How does the least squares method minimize the sum of squared errors?

Tip: Squared errors are always positive, as squaring removes the negative sign, helping to avoid cancellation of positive and negative residuals.