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

Estimated College GPA=0.03+0.9077(High School GPA).

GPAs
College GPA	High School GPA
2.08
2.12
2.27
2.65
3.97
4.09
3.07
3.55
3.46
3.46
3.14
3.76
Step 3 of 3 :  Compute the standard error (se
) of the model. Round your answer to four decimal places.

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

\[
se = \sqrt{\frac{SSE}{n - 2}}
\]

Where:
- \(SSE\) is the sum of squared errors, which measures the variation in the observed values that is not explained by the regression model.
- \(n\) is the number of data points.

### Steps to Compute \(SSE\):

1. **Calculate the predicted college GPA** using the regression equation:
   \[
   \hat{Y}_i = 0.03 + 0.9077 \times X_i
   \]
   where \(X_i\) is the high school GPA for each student.

2. **Compute the residuals** for each data point:
   \[
   e_i = Y_i - \hat{Y}_i
   \]
   where \(Y_i\) is the actual college GPA and \(\hat{Y}_i\) is the predicted college GPA.

3. **Compute the sum of squared errors (SSE)**:
   \[
   SSE = \sum e_i^2
   \]

4. **Calculate the standard error**:
   \[
   se = \sqrt{\frac{SSE}{n - 2}}
   \]

### Step-by-step Calculation:

We have the following data:

| College GPA (\(Y_i\)) | High School GPA (\(X_i\)) |
|-----------------------|---------------------------|
| 2.08                  | 2.12                      |
| 2.27                  | 2.65                      |
| 3.97                  | 4.09                      |
| 3.07                  | 3.55                      |
| 3.46                  | 3.46                      |
| 3.14                  | 3.76                      |

**1. Calculate the predicted college GPAs (\(\hat{Y}_i\))**:

- For \(X_1 = 2.12\): \(\hat{Y}_1 = 0.03 + 0.9077 \times 2.12 = 1.9459\)
- For \(X_2 = 2.65\): \(\hat{Y}_2 = 0.03 + 0.9077 \times 2.65 = 2.9114\)
- For \(X_3 = 4.09\): \(\hat{Y}_3 = 0.03 + 0.9077 \times 4.09 = 3.7215\)
- For \(X_4 = 3.55\): \(\hat{Y}_4 = 0.03 + 0.9077 \times 3.55 = 3.7193\)
- For \(X_5 = 3.46\): \(\hat{Y}_5 = 0.03 + 0.9077 \times 3.46 = 3.6446\)
- For \(X_6 = 3.76\): \(\hat{Y}_6 = 0.03 + 0.9077 \times 3.76 = 3.6339\)

**2. Compute the residuals** (\(e_i = Y_i - \hat{Y}_i\)):

- For \(Y_1 = 2.08\): \(e_1 = 2.08 - 1.9459 = 0.1341\)
- For \(Y_2 = 2.27\): \(e_2 = 2.27 - 2.9114 = -0.6414\)
- For \(Y_3 = 3.97\): \(e_3 = 3.97 - 3.7215 = 0.2485\)
- For \(Y_4 = 3.07\): \(e_4 = 3.07 - 3.7193 = -0.6493\)
- For \(Y_5 = 3.46\): \(e_5 = 3.46 - 3.6446 = -0.1846\)
- For \(Y_6 = 3.14\): \(e_6 = 3.14 - 3.6339 = -0.4939\)

**3. Calculate the sum of squared errors (SSE)**:

\[
SSE = (0.1341)^2 + (-0.6414)^2 + (0.2485)^2 + (-0.6493)^2 + (-0.1846)^2 + (-0.4939)^2
\]
\[
SSE = 0.0179 + 0.4114 + 0.0618 + 0.4216 + 0.0340 + 0.2439 = 1.1906
\]

**4. Compute the standard error**:

The number of data points \(n = 6\). So:

\[
se = \sqrt{\frac{1.1906}{6 - 2}} = \sqrt{\frac{1.1906}{4}} = \sqrt{0.29765} = 0.5462
\]

### Final Answer:

The standard error (se) of the model is **0.5462**.

Would you like a more detailed breakdown of the steps, or do you have any further questions on this topic?

### Related Questions:
1. How do you calculate the coefficient of determination (R²) for this regression model?
2. What is the interpretation of the regression coefficient 0.9077?
3. How do residual plots help in diagnosing regression models?
4. What assumptions are made in simple linear regression?
5. How would the standard error change if we had more data points?

### Tip:
Always visualize your residuals to check if your regression assumptions (like homoscedasticity) are met!

Consider the following data regarding students' college GPAs and high school GPAs. The estimated regression equation is Estimated College GPA = 0.03 + 0.9077(High School GPA). GPAs College GPA High School GPA 2.08 2.12 2.27 2.65 3.97 4.09 3.07 3.55 3.46 3.46 3.14 3.76 Step 3 of 3 : Compute the standard error (se) of the model. Round your answer to four decimal places.

Learn how to compute the standard error (se) for a linear regression model using the formula se = sqrt(SSE / (n - 2)). This step-by-step solution demonstrates the process of calculating predicted values, residuals, and the sum of squared errors (SSE) based on given data. Suitable for students in Grades 10-12.

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