Math Problem Statement

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 2 of 3 : Compute the mean square error (s2e ) for the model. Round your answer to four decimal places.

Solution

To calculate the mean square error (MSEMSE) for the regression model, we need to follow these steps:

Step 1: Calculate the residuals

The residuals (eie_i) are the differences between the observed values of College GPA and the predicted values from the regression equation. The formula for residuals is:

ei=YiY^ie_i = Y_i - \hat{Y}_i

Where:

  • YiY_i is the observed value of College GPA.
  • Y^i\hat{Y}_i is the predicted value of College GPA, which we can calculate using the regression equation:
    Y^i=2.91+0.1998×(High School GPAi)\hat{Y}_i = 2.91 + 0.1998 \times (\text{High School GPA}_i)

Step 2: Compute the residual sum of squares (RSS)

The residual sum of squares is calculated as the sum of squared residuals:

RSS=ei2RSS = \sum e_i^2

Step 3: Calculate the mean square error (MSE)

The mean square error is given by:

MSE=RSSn2MSE = \frac{RSS}{n - 2}

Where:

  • nn is the number of data points (here, n=6n = 6).
  • We subtract 2 from nn because we lose two degrees of freedom due to estimating two parameters (slope and intercept) in the regression model.

Given data:

  • College GPA: [3.13, 2.65, 3.99, 2.37, 3.64, 2.89]
  • High School GPA: [2.65, 2.37, 2.89, 4.51, 2.88, 2.96]

Now, let's calculate the predicted College GPA for each data point, compute the residuals, and finally the mean square error.

I'll proceed with the calculations. numpy as np

Given data

high_school_gpa = np.array([2.65, 2.37, 2.89, 4.51, 2.88, 2.96]) college_gpa = np.array([3.13, 2.65, 3.99, 2.37, 3.64, 2.89])

Regression equation coefficients

intercept = 2.91 slope = 0.1998

Predicted College GPA

predicted_college_gpa = intercept + slope * high_school_gpa

Residuals (difference between observed and predicted College GPA)

residuals = college_gpa - predicted_college_gpa

Residual sum of squares (RSS)

rss = np.sum(residuals**2)

Mean square error (MSE)

n = len(college_gpa) mse = rss / (n - 2)

Round the MSE to 4 decimal places

mse_rounded = round(mse, 4) mse_roundedThe mean square error (MSE) for the regression model is 0.8402 (rounded to four decimal places).

Would you like more details on how this was calculated or have any questions about the process? Here are some related questions that could help expand your understanding:

  1. What is the purpose of the residual sum of squares (RSS) in regression analysis?
  2. How does the degree of freedom (n - 2) affect the MSE calculation?
  3. What other diagnostic measures can be used to evaluate the fit of a regression model?
  4. How would the MSE change if we had more data points?
  5. What does the MSE value tell us about the accuracy of the regression model?

Tip: The closer the MSE is to 0, the better the model fits the data. A high MSE means the model's predictions are farther from the actual values.

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Math Problem Analysis

Mathematical Concepts

Regression Analysis
Residuals
Mean Square Error (MSE)

Formulas

Estimated College GPA = 2.91 + 0.1998 * (High School GPA)
Residuals (e_i) = Y_i - Y_hat_i
Residual Sum of Squares (RSS) = Σ e_i^2
Mean Square Error (MSE) = RSS / (n - 2)

Theorems

Least Squares Regression

Suitable Grade Level

Grades 11-12