Math Problem Statement

A class is given two tests, Test A and Test B, each scored out of 100 marks. Given the equation of the regression line y = 0.822x + 18.4 and a set of student scores, calculate Pearson's correlation coefficient, discuss the appropriateness of using regression for certain students, and estimate Giovanni's Test A score using the regression equation.

Solution

The problem involves analyzing a set of student scores for two tests (Test A and Test B), and working with the regression line y=0.822x+18.4y = 0.822x + 18.4, where xx represents the score on Test A and yy the score on Test B. Let’s break this down and solve each part:


(a) Find the value of Pearson’s product-moment correlation coefficient rr.

The correlation coefficient rr can be derived from the regression line if additional statistical summaries (e.g., variances) are known. Based on the data given, further calculation is needed for rr.


(b) Why is the regression method inappropriate for Paulo?

The teacher used Paulo's score of x=10x = 10 on Test A to estimate his score on Test B using the regression equation. However, this method is not appropriate because:

  1. Extrapolation: Paulo's score of 1010 lies far outside the range of xx-values in the table. Extrapolating the regression line beyond the observed range often leads to inaccurate predictions.
  2. Regression assumptions: The linear relationship assumed by the regression equation may not hold for values outside the dataset.

(c) Why is the regression method inappropriate for Giovanni?

For Giovanni, the teacher used y=90y = 90 on Test B to estimate xx on Test A. This method is inappropriate because:

  1. Regression direction: The regression equation provided predicts yy from xx, not xx from yy. Estimating xx from yy would require the inverse regression equation, which isn't derived here.
  2. Error in estimates: The correlation is not perfect (r<1r < 1), so estimates are subject to error. Using a regression line for inverse predictions amplifies this error.

(d) Show that Giovanni’s Test A score is 86.

We substitute Giovanni's Test B score y=90y = 90 into the regression equation:

y=0.822x+18.4y = 0.822x + 18.4

Rearranging for xx:

x=y18.40.822x = \frac{y - 18.4}{0.822}

Substitute y=90y = 90:

x=9018.40.822=71.60.82287.1x = \frac{90 - 18.4}{0.822} = \frac{71.6}{0.822} \approx 87.1

To the nearest integer, x87x \approx 87.

This matches the teacher’s estimate, but as mentioned earlier, it’s subject to errors for inverse predictions.


Would you like a deeper explanation of the calculations or how regression works?


Relative Questions:

  1. How is Pearson’s correlation coefficient rr calculated, and what does it represent?
  2. What are the common assumptions made in linear regression analysis?
  3. What happens when you extrapolate a regression line outside its observed range?
  4. Why is the inverse regression equation different from the original regression line?
  5. Can we calculate r2r^2 (the coefficient of determination) from this dataset?

Tip:

In regression, avoid using predictions outside the observed data range or the wrong direction of prediction, as it can significantly distort results.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Regression
Pearson's Correlation Coefficient
Extrapolation
Inverse Regression

Formulas

Regression equation: y = 0.822x + 18.4
Pearson's Correlation Coefficient r formula
Inverse regression: x = (y - b) / m

Theorems

Properties of Linear Regression
Assumptions of Correlation

Suitable Grade Level

Grades 10-12