If the correlation coefficient between x and y is 0.5 and the regression coefficient of y on x is 0.5, then the regression coefficient of x on y is:
0.5
The question asks us to find the regression coefficient of x on y, denoted as $b_{xy}$, given the correlation coefficient ($r$) between x and y and the regression coefficient of y on x ($b_{yx}$).
The correlation coefficient ($r$) measures the strength and direction of the linear relationship between two variables. It ranges from -1 to +1.
Regression coefficients measure the average change in the dependent variable for a unit change in the independent variable. $b_{yx}$ represents the regression coefficient of y on x, meaning the change in y per unit change in x. $b_{xy}$ represents the regression coefficient of x on y, meaning the change in x per unit change in y.
There is a fundamental relationship connecting the correlation coefficient and the two regression coefficients:
$\qquad r^2 = b_{yx} \times b_{xy}$
Also, the sign of the correlation coefficient ($r$) must be the same as the sign of both regression coefficients ($b_{yx}$ and $b_{xy}$). Since $r = 0.5$ (positive) and $b_{yx} = 0.5$ (positive), $b_{xy}$ must also be positive.
Therefore, we can write the relationship as:
$\qquad r = \sqrt{b_{yx} \times b_{xy}}$ (when r, $b_{yx}$, and $b_{xy}$ are positive)
Given:
We need to find $b_{xy}$.
Using the formula $r = \sqrt{b_{yx} \times b_{xy}}$:
$\qquad 0.5 = \sqrt{0.5 \times b_{xy}}$
To solve for $b_{xy}$, we square both sides of the equation:
$\qquad (0.5)^2 = (\sqrt{0.5 \times b_{xy}})^2$
$\qquad 0.25 = 0.5 \times b_{xy}$
Now, isolate $b_{xy}$ by dividing both sides by 0.5:
$\qquad b_{xy} = \frac{0.25}{0.5}$
$\qquad b_{xy} = \frac{1/4}{1/2}$
$\qquad b_{xy} = \frac{1}{4} \times \frac{2}{1}$
$\qquad b_{xy} = \frac{2}{4}$
$\qquad b_{xy} = 0.5$
Thus, the regression coefficient of x on y is 0.5.
Let's check if our calculated value of $b_{xy}$ satisfies the original relationship $r^2 = b_{yx} \times b_{xy}$:
Left side: $r^2 = (0.5)^2 = 0.25$
Right side: $b_{yx} \times b_{xy} = 0.5 \times 0.5 = 0.25$
Since Left side = Right side, our calculation is correct.
| Given Value | Symbol | Value |
|---|---|---|
| Correlation Coefficient | $r$ | 0.5 |
| Regression Coefficient of y on x | $b_{yx}$ | 0.5 |
| Regression Coefficient of x on y (Calculated) | $b_{xy}$ | 0.5 |
Given a correlation coefficient of 0.5 and a regression coefficient of y on x of 0.5, the regression coefficient of x on y is calculated to be 0.5 using the relationship between these three statistical measures.
| Concept | Symbol | Description | Key Property |
|---|---|---|---|
| Correlation Coefficient | $r$ | Measures strength and direction of linear relationship (-1 to +1). | $-1 \le r \le 1$ |
| Regression Coefficient of y on x | $b_{yx}$ | Change in y per unit change in x. | Slope of regression line y on x. |
| Regression Coefficient of x on y | $b_{xy}$ | Change in x per unit change in y. | Slope of regression line x on y. |
| Relationship | $r^2 = b_{yx} \times b_{xy}$ | Square of correlation coefficient equals product of regression coefficients. | Sign of r is same as signs of $b_{yx}$ and $b_{xy}$. |
The product of partial regression coefficient \(b_1{_2}{_.3}\) \(b_2{_3}{_.1}\) \(b_3{_1}{_.2}\) equals to:
If six hand-writings were ranked by two judges in a competition and the rankings are as follows:
| Hand Writing | ||||||
| 1 | 2 | 3 | 4 | 5 | 6 | |
| Judge1 | 6 | 5 | 4 | 3 | 2 | 1 |
| Judge2 | 1 | 2 | 3 | 4 | 5 | 6 |
If the difference between the rank of the 4 observations are 2.5, 0.5, -1.5, -1.5, then Spearman's rank correlation coefficient equals to:
For the data set
| x: | 1 | 2 | 3 | 4 |
| y: | 4 | 5 | 3 | 2 |
the regression coefficient b yx (y on x) equals to:
For the two variables X and Y, the following observations are tabulated.
| X: | 3 | 4 | 4 |
| Y: | 10 | 10 | 9 |
The Spearman's correlation coefficient is:
For the variables X and Y, we collect 4 observations with ∑x = 10, ∑y = 14, ∑x 2= 30, ∑y 2= 54, ∑xy = 31. The regression line y on x is
The coefficient of correlations is r between X and Y having standard deviation σx and σy. The tangent of the between two lines of regression is:
In Spearman rank correlation coefficient r s= 1 - \(\dfrac{6∑ d^2}{n(n^2-1)}\) , the maximum value of ∑d 2in case of untied rank is:
If X 1, X 2, … X nis a simple random sample without replacement of size n from a finite population of N units with mean μ and variance σ 2, the covariance of (X i, Xj ) will be:
If x = X- \(\bar{X}\) and y = Y - \(\bar{Y}\) and the number of pairs (X, Y) is n, then the Karl Pearson's coefficient of correlation is: