Math Problem Statement

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 1 of 3 : Compute the sum of squared errors (SSE) for the model. Round your answer to four decimal places

Solution

To calculate the sum of squared errors (SSE) for the regression model, we need to follow these steps:

Step 1: Compute the predicted College GPAs.

The estimated regression equation is:

Y^=4.030.2380X\hat{Y} = 4.03 - 0.2380 \cdot X

Where:

  • Y^\hat{Y} is the predicted College GPA.
  • XX is the High School GPA.

For each pair of High School GPA (XX) and College GPA (YY), we can compute the predicted College GPA (Y^\hat{Y}).

Step 2: Calculate the residuals.

The residuals are the differences between the observed College GPA (YY) and the predicted College GPA (Y^\hat{Y}):

Residual=YY^\text{Residual} = Y - \hat{Y}

Step 3: Square the residuals.

For each pair, square the residuals to get the squared errors:

Squared Error=(YY^)2\text{Squared Error} = (Y - \hat{Y})^2

Step 4: Compute the sum of squared errors (SSE).

Sum all the squared errors:

SSE=i(YiY^i)2SSE = \sum_{i} (Y_i - \hat{Y}_i)^2

Now, let's calculate the predicted College GPAs, the residuals, and the SSE for the given data.

High School GPA (XX)College GPA (YY)Predicted College GPA (Y^\hat{Y})Residual (YY^Y - \hat{Y})Squared Error ((YY^)2(Y - \hat{Y})^2)
4.982.544.030.2380×4.984.03 - 0.2380 \times 4.982.54Y^2.54 - \hat{Y}Residual Squared
3.353.884.030.2380×3.354.03 - 0.2380 \times 3.353.88Y^3.88 - \hat{Y}Residual Squared
3.402.794.030.2380×3.404.03 - 0.2380 \times 3.402.79Y^2.79 - \hat{Y}Residual Squared
2.362.504.030.2380×2.364.03 - 0.2380 \times 2.362.50Y^2.50 - \hat{Y}Residual Squared
3.543.784.030.2380×3.544.03 - 0.2380 \times 3.543.78Y^3.78 - \hat{Y}Residual Squared
2.213.994.030.2380×2.214.03 - 0.2380 \times 2.213.99Y^3.99 - \hat{Y}Residual Squared

Let's now compute the values for each of these steps. Given data high_school_gpa = [4.98, 3.35, 3.40, 2.36, 3.54, 2.21] college_gpa = [2.54, 3.88, 2.79, 2.50, 3.78, 3.99]

Regression equation parameters

intercept = 4.03 slope = -0.2380

Step 1: Compute predicted College GPA (Y_hat) for each data point

predicted_college_gpa = [intercept + slope * x for x in high_school_gpa]

Step 2: Calculate residuals (Y - Y_hat)

residuals = [y - y_hat for y, y_hat in zip(college_gpa, predicted_college_gpa)]

Step 3: Compute squared errors (residuals squared)

squared_errors = [residual ** 2 for residual in residuals]

Step 4: Sum of squared errors (SSE)

sse = sum(squared_errors) sseThe sum of squared errors (SSE) for the regression model is 2.2224 (rounded to four decimal places).

Let me know if you'd like a breakdown of any specific steps or further clarification! Here are some related questions for you:

  1. What does the sum of squared errors (SSE) represent in a regression model?
  2. How does the SSE help assess the quality of the regression model?
  3. What would happen if the SSE were very large in this scenario?
  4. Can the SSE be used to compare different regression models?
  5. How do you compute the total sum of squares (SST) and residual sum of squares (SSR)?

Tip: Always check the residual plots for patterns! If the residuals show a clear pattern (e.g., increasing or decreasing trends), the model might not be fully capturing the underlying data structure.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Regression
Error Calculation
Sum of Squared Errors (SSE)

Formulas

Estimated College GPA = 4.03 + (-0.2380) * High School GPA
Residual = Y - Y_hat
Squared Error = (Y - Y_hat)^2
SSE = Σ (Y_i - Y_hat_i)^2

Theorems

Least Squares Method
Regression Analysis

Suitable Grade Level

Grades 10-12