1. Nov 4, 2007

### steven10137

1. The problem statement, all variables and given/known data
I have have a set of data pairs (x, y);
(1, a)
(2, b)
(3, c)
(4, d)
(5, e)
(6, f)
(7, g)

The least squares regression line for the this set is y=3x-12

Determine the new gradient of this line if the original set of scores has been transformed to;

(6, a+3)
(12, b+3)
(18, c+3)
(24, d+3)
(30, e+3)
(36, f+3)
(42, g+3)

i.e. the x scores have been multiplied by 6, and the +3 has been added to the y scores.
Now from my statistical tables book; I have the formula;
$$m_{gradient} = \frac{{{\rm{covariance}}}}{{{\rm{variance}}}} = \frac{{S_{xy} }}{{S_{x^2 } }}$$

how can I find the new gradient?

$$\begin{array}{l} m_{gradient} = \frac{{{\rm{covariance}}}}{{{\rm{variance}}}} = \frac{{S_{xy} }}{{S_{x^2 } }} = \frac{{ \times 6}}{{ \times 36}} = \times \frac{1}{6} \\ Hence\;gradient\;is\;now\;3 \times \frac{1}{6} = \frac{1}{2} \\ \end{array}$$

I don't really understand how this process works and don't want to assume anything that is wrong

Steven

2. Nov 4, 2007

### HallsofIvy

Staff Emeritus
I'm no expert on statistics but I know that the "gradient" of a line is just its slope. I would have done this, ignoring all the statistical stuff, by arguing that y is now yold+ 3 and x is now 6xold so that yold= y- 3 and xold= x/6. Since you are told that yold= 3xold- 12, you now have y-3= 3(x/6)- 12 or simply y= x/2- 15. The slope (gradient) of that line is 1/2.

3. Nov 4, 2007

### steven10137

Cheers for that HallsofIvy!

seems like the obvious thing to do looking back lol

thanks!

4. Mar 13, 2009