Math Problem Statement

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

Estimated College GPA=4.28+(−0.2134)(High School GPA).

GPAs College GPA High School GPA 3.86 3.15 3.95 2.27 2.86 4.02 3.42 3.08 3.94 4.71 3.00 4.51 Step 1 of 3 : Compute the sum of squared errors (SSE) for the model. Round your answer to four decimal places.

Solution

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

Formula for SSE:

SSE=(yiyi^)2SSE = \sum (y_i - \hat{y_i})^2 Where:

  • yiy_i is the observed value of the dependent variable (College GPA in this case).
  • yi^\hat{y_i} is the predicted value from the regression equation.

Given regression equation:

yi^=4.28+(0.2134)×(High School GPA)\hat{y_i} = 4.28 + (-0.2134) \times (\text{High School GPA})

Now, let's compute the predicted College GPA for each student using the regression equation, and then calculate the error term (difference between actual and predicted values), and finally sum the squared errors.

Step-by-Step Computation:

We'll calculate the predicted College GPA (yi^\hat{y_i}) for each data point and then find the squared error for each.


Given Data:

  1. College GPA: 3.86, High School GPA: 3.15
  2. College GPA: 3.95, High School GPA: 2.27
  3. College GPA: 2.86, High School GPA: 4.02
  4. College GPA: 3.42, High School GPA: 3.08
  5. College GPA: 3.94, High School GPA: 4.71
  6. College GPA: 3.00, High School GPA: 4.51

Let's start by calculating the predicted College GPA for each data point. I'll go ahead and perform the calculations now. numpy as np

Given data

high_school_gpa = np.array([3.15, 2.27, 4.02, 3.08, 4.71, 4.51]) # High School GPA college_gpa = np.array([3.86, 3.95, 2.86, 3.42, 3.94, 3.00]) # College GPA

Regression equation: College GPA = 4.28 + (-0.2134) * High School GPA

predicted_college_gpa = 4.28 + (-0.2134) * high_school_gpa

Compute squared errors (y_i - y_hat_i)^2

squared_errors = (college_gpa - predicted_college_gpa)**2

Compute sum of squared errors (SSE)

sse = np.sum(squared_errors) round(sse, 4) # rounding the result to 4 decimal placesThe sum of squared errors (SSE) for the model is 0.9878.

Would you like further details on the steps or any other questions? Here are some related questions to expand on this:

  1. How can the coefficient of determination (R²) be calculated for this model?
  2. What does the value of SSE indicate about the model’s performance?
  3. How would the SSE change if the regression model were different?
  4. What is the difference between SSE and the residual sum of squares (RSS)?
  5. How can you interpret the slope of the regression equation in the context of this data?

Tip: Always check the assumptions of the regression model, such as linearity, homoscedasticity, and normality of errors, before interpreting the results too deeply.

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Math Problem Analysis

Mathematical Concepts

Linear Regression
Statistical Error Analysis

Formulas

SSE = Σ(y_i - ŷ_i)²
ŷ_i = 4.28 + (-0.2134) * (High School GPA)

Theorems

Regression Analysis
Least Squares Method

Suitable Grade Level

Grades 10-12