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The linear equation $y=2.2+0.6(x+1.2)$ has the slope $0.6$, the given y-intercept $2.2$ and x-intercept $-1.2$. The table is

$$ \begin{array}{c|lcr} x & y \\ \hline 1 & 3.52 \\ 2 & 4.12 \\ 3 & 4.72 \\ 4 & 5.32 \\ 5 & 5.92 \\ \end{array} $$

From the equation (and the blue line on the graph) the actual y-intercept is $2.92$ and the actual x-intercept is $-4.8\bar{6}$ ($-4\frac{13}{15}$).

When writing a regression for this data, how do I relate the "actual" intercepts to the equation input x- and y-intercepts?

Here is what I know.

From the data I can determine the slope $0.6$, and I know that every regression line with slope $0.6$ has to go through the point $(\bar{x}, \bar{y})$ (the means of the x- and y-values) which here is $(3,4.72)$.

The line also has to go through the point $(-b_2, b_0)$, which secretly is $(-1.2, 2.2)$.

The information I actually know when writing a regression from the data alone is

  1. $y=b_0+b_1(x+b_2)$ (linear given)
  2. $b_1 = 0.6$
  3. $\bar{x}=3$
  4. $\bar{y}=4.72$
  5. regression line of slope $b_1=0.6$ has the point $(\bar{x}, \bar{y})=(3,4.72)$
  6. regression line of slope $b_1=0.6$ has the point $(-b_2, b_0)$.

The equations for $b_0$ and $b_2$ are simple but co-dependent:

$$ b_0 = y - b_1(x+b_2) \\ b_2 = \frac{y-b_0}{-b_1}+x $$

Let $b_2=0;x=3$ in $$ b_0=4.72-0.6(3-0) \\ b_0=2.92. $$ Let $b_0=0;x=3$ in $$ b_2 = \frac{4.72-0}{-.6}+3 \\ b_2=-4.8\bar{6}. $$


These are the observed x- and y-intercepts, but not the orginal inputs.

Put these back into the original linear equation where $x=3$:
$$4.72=2.92+0.6(3+-4.8\bar{6})$$

I find that $4.72=2.92+4.72$ thus $4.72=7.64$.


Let $x=3$ in 2-variable derivation of $b_0$ where $y=2.2+0.6(x)$ (1 unknown):

$$ \bar{y} = b_0+0.6(\bar{x}) \\ 4 = b_0+1.8 \\ 2.2 = b_0. $$ This is correct.

If I let $x=3$ and use the same strategy for 3-variable derivation of $b_0$ where $y=2.2+0.6(x+1.2)$ (2 unknowns):

$$ \bar{y} = b_0+0.6(\bar{x}+b_2) \\ 4.72 = b_0+1.8+.6b_2 \\ 2.92 = b_0+0.6b_2. $$

The result is balanced:

$$ 2.92 = 2.2+0.6(1.2) \\ 2.92 = 2.2+0.72 \\ 2.92 = 2.92. $$

but does not give $b_0$ or $b_2$ because they are defined in terms of one another. There are infinitely many real solutions to $2.92 = b_0+0.6b_2$.

How shall I decouple the two intercepts to find the original equation $y=2.2+0.6(x+1.2)$ from the data and knowing only the 6 listed facts above?

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1 Answer 1

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If $\bar{x} = 3$, and $\bar{y} = 4.72$ and the slope is $0.6$ then the x-intercept will be:

$3 - (\frac{4.72}{0.6}) = -4.8\bar{6}$

The y-intercept will be:

$4.72 - (3\cdot 0.6) = 2.92$

The equation is $y = 0.6x + 2.92$

In the form you wrote, $y=2.2+0.6(x+1.2)$ which is a modified point slope form of the equation $(y - y1 = m(x - x1))$ whereby $(-1.2, 2.2)$ is a point and not an x or y intercept.

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