The equations of two regression lines are \(10x-4y=80\) and \(10y-9x=-40\). Find(a) \(\bar{x}\) and \(\bar{y}\)(b) \(b_{YX}\) and \(b_{XY}\)(c) r(d) If var(Y) = 36, obtain var (X).

Answer: \(\bar{x}=10,\ \bar{y}=5;\ b_{YX}=0.9,\ b_{XY}=0.4;\ r=0.6;\ \text{var}(X)=16\)

(a) The means satisfy both equations: \(10\bar{x}-4\bar{y}=80\) and \(-9\bar{x}+10\bar{y}=-40\). Multiplying the first by 5 and the second by 2 and adding: \(50\bar{x}-18\bar{x}=400-80\Rightarrow32\bar{x}=320\Rightarrow\bar{x}=10\). Then \(100-4\bar{y}=80\Rightarrow\bar{y}=5\).

(b) Take \(10x-4y=80\) as the line of X on Y: \(x=0.4y+8\Rightarrow b_{XY}=0.4\). Take \(10y-9x=-40\) as the line of Y on X: \(y=0.9x-4\Rightarrow b_{YX}=0.9\). Since \(b_{YX}\cdot b_{XY}=0.36\le1\), this choice is correct. (The other choice gives \(\frac{10}{4}\times\frac{10}{9}\gt1\), which is impossible.)

(c) \(r^2=b_{YX}\cdot b_{XY}=0.36\Rightarrow r=0.6\) (positive, as both coefficients are positive).

(d) \(\sigma_Y=6\). \(b_{XY}=r\frac{\sigma_X}{\sigma_Y}\Rightarrow0.4=0.6\times\frac{\sigma_X}{6}\Rightarrow\sigma_X=4\). Hence \(\text{var}(X)=16\).