Consider the following data regarding students' college GPAs and high school GPAs. The estimated regression equation is

Estimated College GPA=4.03+(−0.2380)(High School GPA).

GPAs
College GPA	High School GPA
2.54
4.98
3.88
3.35
2.79
3.40
2.50
2.36
3.78
3.54
3.99
2.21
Step 3 of 3 :  Compute the standard error (se
) of the model. Round your answer to four decimal places.

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.

Consider the following data regarding students' college GPAs and high school GPAs. The estimated regression equation is Estimated College GPA=4.03+(−0.2380)(High School GPA). GPAs College GPA High School GPA 2.54 4.98 3.88 3.35 2.79 3.40 2.50 2.36 3.78 3.54 3.99 2.21 Step 3 of 3 : Compute the standard error (se) of the model. Round your answer to four decimal places.

Learn how to calculate the standard error of a linear regression model with high school GPAs and college GPAs. The calculation steps and a detailed explanation of the standard error formula are provided, using an example with 6 data points. Suitable for students in Grades 11-12.

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