Understanding the Mean Values and Correlation Coefficient from Regression Lines

Regression analysis is a fundamental statistical method used to understand the relationship between variables. This article dives into how to derive the mean values and correlation coefficient when given the regression lines of two random variables. We will solve a problem involving the equations 3x2y 26 and 6x y 31 to illustrate this process.

Step-by-Step Guide to Deriving Mean Values and Correlation Coefficient

Given the regression lines:

Rearranging the Regression Equations

First Regression Line: 3x 2y 26 Second Regression Line: 6x y 31

We rearrange these into the slope-intercept form ( y mx b ).

First Regression Line: 2y 26 - 3x

( y -dfrac{3}{2}x 13 )

Second Regression Line: y -6x 31

Step 1: Finding the Means

The means of X and Y can be derived from the intersection of the two regression lines.

Setting Equations Equal to Find Intersection:

First Equation: ( y -dfrac{3}{2}x 13 )

Second Equation: ( y -6x 31 )

Solving these two equations:

Setting the Equations Equal:

( -dfrac{3}{2}x 13 -6x 31 )

Solving for x:

( -dfrac{3}{2}x 6x 31 - 13 )

( dfrac{9}{2}x 18 )

( x 4 )

Finding y:

Substituting ( x 4 ) into the first regression line:

( y -dfrac{3}{2}(4) 13 -6 13 7 )

Therefore, the means are:

( bar{x} 4 )

( bar{y} 7 )

Step 3: Finding the Correlation Coefficient

The correlation coefficient ( r ) can be calculated using the slopes of the regression lines. The relationship is given by:

( r sqrt{m_{YX} cdot m_{XY}} )

Where:

( m_{YX} -dfrac{3}{2} ) - slope of Y on X ( m_{XY} -dfrac{1}{6} ) - slope of X on Y

Calculating ( r ):

( r sqrt{left(-dfrac{3}{2}right) cdot left(-dfrac{1}{6}right)} sqrt{dfrac{3}{12}} sqrt{dfrac{1}{4}} dfrac{1}{2} )

Summary of Results:

Mean values: ( bar{x} 4 ) Mean values: ( bar{y} 7 ) Correlation coefficient: ( r dfrac{1}{2} )

The means are 4 for X and 7 for Y, and the correlation coefficient is 0.5.

Advanced Insights

The given regression lines are 3x2y 26 and 6x y 31. The mean value is always on the regression lines. Therefore, the point of intersection common to the two regression lines gives the mean value (x, y). The intersection point is determined to be (4, 7), and the mean values are x' 4 and y' 7.

For further clarity:

First Line as Regression Line of x on y:

( 3x -2y 26 )

( x -dfrac{2}{3}y dfrac{26}{3} )

Regression coefficient of x on y is ( b_{yx} -dfrac{2}{3} )

Second Line as Regression Line of x on y:

( 6x -y 31 )

( x -dfrac{1}{6}y dfrac{31}{6} )

Regression coefficient of x on y is ( b_{yx} -dfrac{1}{6} )

Using the Second Line as Regression Line of y on x:

( text{Regression coefficient of y on x is } b_{xy} -dfrac{3}{2} )

( r b_{xy} cdot b_{yx} -dfrac{3}{2} cdot -dfrac{1}{6} dfrac{1}{4} )

( r pm dfrac{1}{2} )

Since the regression coefficients are negative, the correlation coefficient ( r ) is negative.

( r -dfrac{1}{2} )