Question

The regression lines of a sample are $x + 6y = 6$ and $3x + 2y = 10$. Then the sample means and the correlation coefficient are

• A. $\bar{x} = 3, \bar{y} = \frac{1}{2}, r = -\frac{1}{3}$
• B. $\bar{x} = 3, \bar{y} = \frac{1}{2}, r = \frac{1}{3}$
• C. $\bar{x} = \frac{1}{2}, \bar{y} = 3, r = -\frac{1}{3}$
• D. $\bar{x} = \frac{1}{2}, \bar{y} = 3, r = \frac{1}{3}$

Hint:

If your assumption for the regression lines results in $|r| > 1$, simply swap the assignments of the lines.
The sign of the correlation coefficient $r$ is always identical to the sign of the slopes in the regression equations.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question asks us to find the sample means ($\bar{x}, \bar{y}$) and the correlation coefficient ($r$) given the equations of two linear regression lines.

Step 2: Key Formula or Approach:
1. The two regression lines always intersect at the point representing the sample means $(\bar{x}, \bar{y})$.
2. If the regression line of $y$ on $x$ has slope $b_{yx}$ and the regression line of $x$ on $y$ has slope $b_{xy}$, then the correlation coefficient $r$ is:
\[ r = \pm \sqrt{b_{yx} \cdot b_{xy}} \] The sign of $r$ must match the signs of both $b_{yx}$ and $b_{xy}$ (since both slopes must have the same sign). Additionally, the correlation coefficient must satisfy $|r| \leq 1$.

Step 3: Detailed Explanation:

• First, we find the sample means $(\bar{x}, \bar{y})$ by solving the two regression equations simultaneously:
\[ \text{Equation 1: } x + 6y = 6 \implies x = 6 - 6y \]
\[ \text{Equation 2: } 3x + 2y = 10 \]

• Substitute the expression for $x$ from Equation 1 into Equation 2:
\[ 3(6 - 6y) + 2y = 10 \]
\[ 18 - 18y + 2y = 10 \]
\[ 18 - 16y = 10 \implies 16y = 8 \implies y = \frac{1}{2} \]

• Substitute $y = \frac{1}{2}$ back into the equation for $x$:
\[ x = 6 - 6\left(\frac{1}{2}\right) = 6 - 3 = 3 \]

• Thus, the sample means are:
\[ \bar{x} = 3, \quad \bar{y} = \frac{1}{2} \]

• Next, we determine the correlation coefficient $r$. We must assign one line as the regression of $y$ on $x$ and the other as $x$ on $y$ such that the product of their slopes satisfies $|r| \leq 1$.

• Let us assume Equation 1 is the regression line of $y$ on $x$:
\[ 6y = -x + 6 \implies y = -\frac{1}{6}x + 1 \implies b_{yx} = -\frac{1}{6} \]

• Let us assume Equation 2 is the regression line of $x$ on $y$:
\[ 3x = -2y + 10 \implies x = -\frac{2}{3}y + \frac{10}{3} \implies b_{xy} = -\frac{2}{3} \]

• Now, we calculate $r^2$:
\[ r^2 = b_{yx} \cdot b_{xy} = \left(-\frac{1}{6}\right) \cdot \left(-\frac{2}{3}\right) = \frac{2}{18} = \frac{1}{9} \]

• Taking the square root:
\[ |r| = \sqrt{\frac{1}{9}} = \frac{1}{3} \]

• Since this value is less than 1 ($|r| = 0.33 \leq 1$), our assignment is correct.

• Because both regression coefficients $b_{yx}$ and $b_{xy}$ are negative, the correlation coefficient $r$ must also be negative:
\[ r = -\frac{1}{3} \]

Step 4: Final Answer
The means are $\bar{x} = 3, \bar{y} = \frac{1}{2}$ and the correlation coefficient is $r = -\frac{1}{3}$, which corresponds to option (A).