Math Problem Statement

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Find the mean values of the variables X and Y and the correlation coefficient for the following regression equations: 2Y - X - 50 = 0; 3Y - 2X - 10 = 0.

Solution

To solve this problem, we need to handle two tasks:

  1. Find the mean values of X and Y from the given regression equations.
  2. Find the correlation coefficient based on the regression equations.

The given regression equations are:

  1. 2YX50=02Y - X - 50 = 0
  2. 3Y2X10=03Y - 2X - 10 = 0

Step 1: Solve the system of equations

We solve these equations simultaneously to find the values of Xˉ\bar{X} and Yˉ\bar{Y} (the mean values of XX and YY).

Equation 1:

2YX50=0X=2Y502Y - X - 50 = 0 \quad \Rightarrow \quad X = 2Y - 50

Equation 2:

3Y2X10=03Y - 2X - 10 = 0

Substitute X=2Y50X = 2Y - 50 into the second equation: 3Y2(2Y50)10=03Y - 2(2Y - 50) - 10 = 0

Simplifying: 3Y4Y+10010=0Y+90=03Y - 4Y + 100 - 10 = 0 \quad \Rightarrow \quad -Y + 90 = 0

Y=90Y = 90

Now that we have Y=90Y = 90, substitute this value into the first equation to find XX: X=2(90)50=18050=130X = 2(90) - 50 = 180 - 50 = 130

Thus, the mean values are: Xˉ=130,Yˉ=90\bar{X} = 130, \quad \bar{Y} = 90

Step 2: Find the Correlation Coefficient

The correlation coefficient rr is related to the regression coefficients. The general regression equations for two variables XX and YY are: XXˉ=rσXσY(YYˉ)X - \bar{X} = r \cdot \frac{\sigma_X}{\sigma_Y} \cdot (Y - \bar{Y}) YYˉ=rσYσX(XXˉ)Y - \bar{Y} = r \cdot \frac{\sigma_Y}{\sigma_X} \cdot (X - \bar{X}) where σX\sigma_X and σY\sigma_Y are the standard deviations of XX and YY, respectively.

From the given regression equations, we can extract the regression coefficients:

  • The regression equation of XX on YY is X=2Y50X = 2Y - 50, so the regression coefficient bXY=2b_{XY} = 2.
  • The regression equation of YY on XX is Y=23X+103Y = \frac{2}{3}X + \frac{10}{3}, so the regression coefficient bYX=23b_{YX} = \frac{2}{3}.

The correlation coefficient rr is the geometric mean of the regression coefficients: r=bXYbYX=223=43=231.155r = \sqrt{b_{XY} \cdot b_{YX}} = \sqrt{2 \cdot \frac{2}{3}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}} \approx 1.155

Final Answer:

  • Mean of XX: 130130
  • Mean of YY: 9090
  • Correlation coefficient rr: 231.155\frac{2}{\sqrt{3}} \approx 1.155

Would you like any further details or clarifications?


Related questions:

  1. What are the assumptions behind calculating the correlation coefficient?
  2. How can we interpret the value of the correlation coefficient in this context?
  3. How would the solution change if the regression equations were different?
  4. What is the geometric interpretation of the correlation coefficient?
  5. Can the correlation coefficient ever be greater than 1?

Tip: When solving regression-based problems, always check if the system of equations is consistent and leads to a meaningful solution.

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Math Problem Analysis

Mathematical Concepts

Regression Analysis
Algebra
Correlation Coefficient

Formulas

Regression Equation: X = bYX * Y + c
Regression Equation: Y = bXY * X + d
Correlation Coefficient: r = sqrt(bXY * bYX)

Theorems

Correlation Coefficient Theorem
Simultaneous Equations Solving

Suitable Grade Level

Undergraduate Level (Statistics/Regression)