Q.Using $\bar x=3$, $\bar y=4$, $b_{yx}=0.6$ and $b_{xy}=1.0$ from Worked Example 4, write down the two regression line equations and verify that they intersect at $(\bar x,\bar y)$.

Substitute the known means and coefficients into the two regression-line formulas, then solve them together to check where they cross.

Step 1 — Write the regression line of $y$ on $x$. $y-\bar y = b_{yx}(x-\bar x) \Rightarrow y-4 = 0.6(x-3) \Rightarrow y = 0.6x-1.8+4 = 0.6x+2.2$.

Step 2 — Write the regression line of $x$ on $y$. $x-\bar x = b_{xy}(y-\bar y) \Rightarrow x-3 = 1.0(y-4) \Rightarrow x=y-1$, which rearranges to $y=x+1$.

Step 3 — Solve the two equations simultaneously. Setting the two expressions for $y$ equal: $0.6x+2.2 = x+1 \Rightarrow 2.2-1 = x-0.6x \Rightarrow 1.2=0.4x \Rightarrow x=3$. Substituting back: $y=x+1=3+1=4$.

Step 4 — Confirm this matches the known means. The intersection point, $(3,4)$, is exactly $(\bar x,\bar y)=(3,4)$ — confirming, as the theory predicts, that both regression lines pass through the point of means. …

The zero-substitution check in the Independent check above (plugging $x=\bar x,y=\bar y$ directly into the un-rearranged deviation-form equations) is faster than solving the system from scratch, and works for an …

Forgetting to rearrange $x-\bar x=b_{xy}(y-\bar y)$ into a $y=\ldots$ form before equating it with the other line (or equating the two equations in inconsis …