Using the regression line equation to predict y when x = 2.5

x_value = 2.5 y_estimated = b0 + b1 * x_value

Rounding the result to three decimal places

y_estimated_rounded = round(y_estimated, 3) y_estimated_roundedThe estimated value of $y$ (midterm grade) when $x = 2.5$ (hours spent studying) is approximately $80.958$, rounded to three decimal places.

Would you like more details or have any questions?

Related Questions:

1. How would the estimate change for different values of $x$?
2. What is the margin of error in this estimate?
3. How can residuals be calculated for this prediction?
4. How would you interpret the accuracy of this prediction?
5. Can you predict the grade if the student studies 6 hours?
6. How would the regression line change if the sample size increases?
7. Can you use this model for values of $x$ outside the observed range?
8. What is the confidence interval for this prediction?

Tip:

To ensure accuracy, predictions outside the observed data range (extrapolation) should be approached with caution, as the model may not hold beyond the original data.