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

Estimated College GPA=2.91+0.1998(High School GPA).

GPAs
College GPA	High School GPA
3.13
2.65
3.99
2.37
3.64
2.89
3.98
4.51
3.34
2.88
3.04
2.96
Step 3 of 3 :  Compute the standard error (se
) of the model. Round your answer to four decimal places.

To compute the standard error of the model \( s_e \), we will follow these steps:

### Step 1: Find the residuals for each data point.
The residual for each data point is the difference between the actual college GPA and the predicted college GPA based on the regression equation.

The regression equation is:

\[
\hat{Y} = 2.91 + 0.1998 \times X
\]

where:
- \( \hat{Y} \) is the predicted College GPA,
- \( X \) is the High School GPA.

### Step 2: Compute the residuals for each observation.

For each data point:

\[
\text{Residual} = \text{Actual College GPA} - \hat{\text{College GPA}}
\]

We'll compute the predicted College GPAs for the given High School GPAs using the equation above and subtract the actual College GPAs.

### Step 3: Compute the standard error \( s_e \).

The standard error of the estimate is calculated using the formula:

\[
s_e = \sqrt{\frac{\sum_{i=1}^{n} (Y_i - \hat{Y}_i)^2}{n - 2}}
\]

where:
- \( Y_i \) is the actual College GPA,
- \( \hat{Y}_i \) is the predicted College GPA from the regression equation,
- \( n \) is the number of data points (in this case, 6 data points).

Let me calculate that for you now.

### Data:

| College GPA (Y) | High School GPA (X) | Predicted College GPA \( \hat{Y} \) | Residuals \( Y - \hat{Y} \) |
|-----------------|---------------------|-----------------------------------|----------------------------|
| 3.13            | 2.65                | \( 2.91 + 0.1998 \times 2.65 = 3.44 \)  | \( 3.13 - 3.44 = -0.31 \)  |
| 3.99            | 2.37                | \( 2.91 + 0.1998 \times 2.37 = 3.30 \)  | \( 3.99 - 3.30 = 0.69 \)   |
| 3.64            | 2.89                | \( 2.91 + 0.1998 \times 2.89 = 3.47 \)  | \( 3.64 - 3.47 = 0.17 \)   |
| 3.98            | 4.51                | \( 2.91 + 0.1998 \times 4.51 = 3.79 \)  | \( 3.98 - 3.79 = 0.19 \)   |
| 3.34            | 2.88                | \( 2.91 + 0.1998 \times 2.88 = 3.47 \)  | \( 3.34 - 3.47 = -0.13 \)  |
| 3.04            | 2.96                | \( 2.91 + 0.1998 \times 2.96 = 3.47 \)  | \( 3.04 - 3.47 = -0.43 \)  |

### Step 4: Square the residuals, sum them, and calculate the standard error.

\[
\sum \text{Residuals}^2 = (-0.31)^2 + (0.69)^2 + (0.17)^2 + (0.19)^2 + (-0.13)^2 + (-0.43)^2
\]
\[
\sum \text{Residuals}^2 = 0.0961 + 0.4761 + 0.0289 + 0.0361 + 0.0169 + 0.1849 = 0.838
\]

Now, we compute the standard error:

\[
s_e = \sqrt{\frac{0.838}{6 - 2}} = \sqrt{\frac{0.838}{4}} = \sqrt{0.2095} \approx 0.4577
\]

### Final Answer:

The standard error of the model \( s_e \) is approximately **0.4577**.

---

Would you like any further clarification on this process, or any additional steps explained? Here's a set of related questions for you:

1. What does the standard error \( s_e \) indicate in regression analysis?
2. How would the standard error change if the number of data points increased?
3. How can you calculate the coefficient of determination (R²) from the regression equation?
4. What is the importance of residuals in evaluating regression models?
5. How can you use the standard error to assess the reliability of predictions from the model?

Tip: When interpreting the standard error in regression, smaller values suggest that the model’s predictions are closer to the actual data points, indicating a better fit.

Consider the following data regarding students' college GPAs and high school GPAs. The estimated regression equation is Estimated College GPA = 2.91 + 0.1998(High School GPA). GPAs College GPA High School GPA 3.13 2.65 3.99 2.37 3.64 2.89 3.98 4.51 3.34 2.88 3.04 2.96 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 (s_e) for a regression model predicting college GPAs from high school GPAs. This step-by-step solution covers residuals, regression equations, and standard error formula with real data. Suitable for students in Grades 11-12.

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