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Question: When \(r=-1\), the lines of regression will be _______. (A) Negative slope (B) Positive slope ...

When r=−1r=-1, the lines of regression will be _______.
(A) Negative slope
(B) Positive slope
(C) Running downwards from left to right
(D) Both (A) and (C)

Explanation

Solution

We solve this question by first considering the formula for regression lines of y on x and x on y, (y−y‾)=rSySx(x−x‾)\left( y-\overline{y} \right)=r\dfrac{{{S}_{y}}}{{{S}_{x}}}\left( x-\overline{x} \right) and (x−x‾)=rSxSy(y−y‾)\left( x-\overline{x} \right)=r\dfrac{{{S}_{x}}}{{{S}_{y}}}\left( y-\overline{y} \right) respectively. Then we substitute the value r=−1r=-1 in them and find their slopes by comparing them with the general form of line, y=mx+cy=mx+c. Then we consider the property that standard deviation is always greater than or equal to zero and find whether the slope is positive or negative.

Complete step by step answer:
First let us consider the regression lines.
There are two types of regression lines

  1. Regression line of y on x
  2. Regression line of x on y
    The regression line of y on x is given as
    ⇒(y−y‾)=rSySx(x−x‾)\Rightarrow \left( y-\overline{y} \right)=r\dfrac{{{S}_{y}}}{{{S}_{x}}}\left( x-\overline{x} \right)
    Similarly, we can give the regression line of x on y as,
    ⇒(x−x‾)=rSxSy(y−y‾)\Rightarrow \left( x-\overline{x} \right)=r\dfrac{{{S}_{x}}}{{{S}_{y}}}\left( y-\overline{y} \right)
    Where Sy={{S}_{y}}= Standard deviation of y
    Sx={{S}_{x}}= Standard deviation of x
    r=r= Correlation co-efficient
    We are given that r=−1r=-1. Then the lines will become,
    The regression line of y on x becomes
    ⇒(y−y‾)=(−1)SySx(x−x‾) ⇒(y−y‾)=−SySx(x−x‾) \begin{aligned} & \Rightarrow \left( y-\overline{y} \right)=\left( -1 \right)\dfrac{{{S}_{y}}}{{{S}_{x}}}\left( x-\overline{x} \right) \\\ & \Rightarrow \left( y-\overline{y} \right)=-\dfrac{{{S}_{y}}}{{{S}_{x}}}\left( x-\overline{x} \right) \\\ \end{aligned}
    Now let us consider the general form of equation of a line, y=mx+cy=mx+c. Comparing the regression line with this formula we get the slope of the regression line as,
    ⇒Slope=−SySx..........(1)\Rightarrow Slope=-\dfrac{{{S}_{y}}}{{{S}_{x}}}..........\left( 1 \right)
    Similarly, the regression line of x on y becomes,
    ⇒(x−x‾)=(−1)SxSy(y−y‾) ⇒(x−x‾)=−SxSy(y−y‾) ⇒(y−y‾)=−SySx(x−x‾) \begin{aligned} & \Rightarrow \left( x-\overline{x} \right)=\left( -1 \right)\dfrac{{{S}_{x}}}{{{S}_{y}}}\left( y-\overline{y} \right) \\\ & \Rightarrow \left( x-\overline{x} \right)=-\dfrac{{{S}_{x}}}{{{S}_{y}}}\left( y-\overline{y} \right) \\\ & \Rightarrow \left( y-\overline{y} \right)=-\dfrac{{{S}_{y}}}{{{S}_{x}}}\left( x-\overline{x} \right) \\\ \end{aligned}
    Now let us consider the general form of equation of a line, y=mx+cy=mx+c. Comparing the regression line with this formula we get the slope of the regression line as,
    ⇒Slope=−SySx...........(2)\Rightarrow Slope=-\dfrac{{{S}_{y}}}{{{S}_{x}}}...........\left( 2 \right)
    From equation (1) and (2) we get the slopes of regression lines are same when r=−1r=-1, that is,
    ⇒Slope=−SySx\Rightarrow Slope=-\dfrac{{{S}_{y}}}{{{S}_{x}}}
    Now let us consider the property of standard deviation. Standard deviation is always greater than or equal to zero.
    So, we can say that,
& \Rightarrow \dfrac{{{S}_{y}}}{{{S}_{x}}}>0 \\\ & \Rightarrow -\dfrac{{{S}_{y}}}{{{S}_{x}}}<0 \\\ \end{aligned}$$ So, we get that, $\therefore Slope=-\dfrac{{{S}_{y}}}{{{S}_{x}}}<0$ So, we get that when $r=-1$, the regression lines will have negative slope. As the slope is negative the line runs downward from left to right. **Hence the answer is Option D.** **Note:** The common mistake one makes while solving this question is one might solve the question till finding the slope and mark the answer as Option A without checking the possibility of another correct option here, that is the regression line runs downward from left to right.