To compute the standard error ($s_e$) of the regression model, we use the following formula:

$s_e = \sqrt{\frac{ \sum{(y_i - \hat{y}_i)^2} }{n - 2}}$

Where:

• $y_i$ is the actual value of the College GPA,
• $\hat{y}_i$ is the predicted College GPA from the regression equation,
• $n$ is the number of data points (in this case, $n = 6$ because there are 6 data points).

Steps:

1. Find the predicted College GPA ($\hat{y}_i$) for each student using the given regression equation:
   $\hat{y}_i = 4.03 + (-0.2380)(\text{High School GPA})$
2. Calculate the squared residuals: $(y_i - \hat{y}_i)^2$
3. Sum the squared residuals.
4. Substitute the values into the standard error formula to find $s_e$.

Let me compute that for you.

First, I'll calculate the predicted College GPA for each student.

Data provided:

College GPA ($y_i$)	High School GPA	Predicted College GPA ($\hat{y}_i$)
2.54	4.98	$\hat{y}_1 = 4.03 + (-0.238)(4.98)$
3.88	3.35	$\hat{y}_2 = 4.03 + (-0.238)(3.35)$
2.79	3.40	$\hat{y}_3 = 4.03 + (-0.238)(3.40)$
2.50	2.36	$\hat{y}_4 = 4.03 + (-0.238)(2.36)$
3.78	3.54	$\hat{y}_5 = 4.03 + (-0.238)(3.54)$
3.99	2.21	$\hat{y}_6 = 4.03 + (-0.238)(2.21)$

Let's calculate the predicted values and proceed with the error computation. numpy as np

Given data

high_school_gpa = np.array([4.98, 3.35, 3.40, 2.36, 3.54, 2.21]) college_gpa = np.array([2.54, 3.88, 2.79, 2.50, 3.78, 3.99])

Regression equation constants

intercept = 4.03 slope = -0.238

Predicted College GPA (y_hat) for each data point

predicted_gpa = intercept + slope * high_school_gpa

Calculate the residuals (actual - predicted)

residuals = college_gpa - predicted_gpa

Calculate the squared residuals

squared_residuals = residuals**2

Sum of squared residuals

sum_squared_residuals = np.sum(squared_residuals)

Number of data points

n = len(college_gpa)

Standard error (se) formula

standard_error = np.sqrt(sum_squared_residuals / (n - 2))

Round to four decimal places

standard_error_rounded = round(standard_error, 4)

standard_error_roundedThe standard error ($s_e$) of the model is approximately 0.7454.

Would you like more details on the calculation steps or have any further questions? Here are some related questions that might interest you:

1. How do changes in the high school GPA affect the college GPA based on this model?
2. What is the interpretation of the slope and intercept in this regression equation?
3. How can we use the standard error to assess the accuracy of the regression model?
4. What would happen to the standard error if more data points were added to the model?
5. How do you calculate the $R^2$ value for this regression?

Tip: The standard error helps in understanding how well the regression model fits the data. A lower standard error indicates that the model's predictions are close to the actual values.