Question 1196666
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The regression line is of the form y = mx+b
m = slope
b = y intercept


The slope formula for a linear regression is this particularly complex equation
*[tex \Large m = \frac{n*\left(\sum \text{x}y\right)-\left(\sum \text{x}\right)*\left(\sum y\right)}{n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2}]


We don't need to calculate the value of m itself. We just need to see how m changes.
The x values represent the total length in cm
To go from cm to inches, we multiply by 0.3937 since 1 cm = 0.3937 inches approximately


This means each x updates to  0.3937x


Let's replace each x with 0.3937x and do a bit of algebra like so:
*[tex \Large m = \frac{n*\left(\sum \text{x}y\right)-\left(\sum \text{x}\right)*\left(\sum y\right)}{n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2}]


*[tex \Large m = \frac{n*\left(\sum 0.3937\text{x}y\right)-\left(\sum 0.3937\text{x}\right)*\left(\sum y\right)}{n*\left(\sum (0.3937\text{x})^2\right)-\left(\sum 0.3937\text{x}\right)^2}]


*[tex \Large m = \frac{0.3937n*\left(\sum \text{x}y\right)-0.3937\left(\sum \text{x}\right)*\left(\sum y\right)}{0.3937^2n*\left(\sum \text{x}^2\right)-0.3937^2\left(\sum \text{x}\right)^2}]


*[tex \Large m = \frac{0.3937\left(n*\left(\sum \text{x}y\right)-\left(\sum \text{x}\right)*\left(\sum y\right)\right)}{0.3937^2\left(n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2\right)}]


*[tex \Large m = \frac{n*\left(\sum \text{x}y\right)-\left(\sum \text{x}\right)*\left(\sum y\right)}{0.3937\left(n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2\right)}]


This shows that
{{{newSlope = (1/(0.3937))*(oldSlope)}}}
or
{{{newSlope = (oldSlope)/(0.3937)}}}
and this occurs after we scaled the x values to get 0.3937x (i.e multiplied each x value by 0.3937 to convert from cm to inches)


Through similar steps we can replace each y with 2.2y and have this:
*[tex \Large m = \frac{n*\left(\sum \text{x}y\right)-\left(\sum \text{x}\right)*\left(\sum y\right)}{n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2}]


*[tex \Large m = \frac{n*\left(\sum \text{x}*2.2y\right)-\left(\sum \text{x}\right)*\left(\sum 2.2y\right)}{n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2}]


*[tex \Large m = \frac{2.2n*\left(\sum \text{x}y\right)-2.2\left(\sum \text{x}\right)*\left(\sum y\right)}{n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2}]


*[tex \Large m = \frac{2.2\left(n*\left(\sum \text{x}y\right)-\left(\sum \text{x}\right)*\left(\sum y\right)\right)}{n*\left(\sum \text{x}^2\right)-\left(\sum \text{x}\right)^2}]


So we arrive at the fact {{{newSlope = 2.2*(oldSlope)}}} when we scaled each y value by 2.2 (i.e. multiplied each weight by 2.2 to convert from kg to pounds)


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Put the two ideas together and we have
{{{newSlope = (2.2/0.3937)*(oldSlope)}}}


{{{newSlope = 5.588*(oldSlope)}}}
The value 5.588 is approximate


You can think of it like this:
slope = rise/run
slope = (change in y)/(change in x)
new slope = (2.2*(change in y))/(0.3937*(change in x))
new slope = (2.2/0.3937)*( (change in y)/(change in x) )
new slope = (2.2/0.3937)*(old slope)
<font color=red>new slope = 5.588*(old slope)</font>


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Answer: <font color=red>Increase by a factor of 5.588
Since 5.588 = 2.2/0.3937</font>


Note: you likely wont need to write down those complicated formulas for your homework. Those are for illustrative purposes mostly. 
You could use the much faster method that the tutor @ewatrrr talked about.
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