Finding the New Gradient: A Statistical Tables Book Guide

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Homework Help Overview

The discussion revolves around a problem involving the transformation of a set of data pairs and the calculation of the new gradient of a least squares regression line. The original poster presents a set of data points and a regression line equation, seeking clarification on how to determine the new gradient after applying specific transformations to the data.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between the original and transformed data, questioning how the transformations affect the gradient. One participant suggests a more intuitive approach by rewriting the equations based on the transformations applied to the x and y values.

Discussion Status

The discussion is ongoing, with some participants providing insights and alternative methods for understanding the problem. There is no explicit consensus on the best approach, but participants are engaging with the concepts and clarifying their understanding of the transformations involved.

Contextual Notes

There is a request for the source of the formula used for calculating the gradient and correlation, indicating a potential gap in the original poster's resources or understanding of the statistical concepts involved.

steven10137
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Homework Statement


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;
[tex]m_{gradient} = \frac{{{\rm{covariance}}}}{{{\rm{variance}}}} = \frac{{S_{xy} }}{{S_{x^2 } }}[/tex]

how can I find the new gradient?

The answer says;
[tex]\begin{array}{l}<br /> m_{gradient} = \frac{{{\rm{covariance}}}}{{{\rm{variance}}}} = \frac{{S_{xy} }}{{S_{x^2 } }} = \frac{{ \times 6}}{{ \times 36}} = \times \frac{1}{6} \\ <br /> Hence\;gradient\;is\;now\;3 \times \frac{1}{6} = \frac{1}{2} \\ <br /> \end{array}[/tex]

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

thanks in advance
Steven
 
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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.
 
Cheers for that HallsofIvy!

seems like the obvious thing to do looking back lol

thanks!
 
hi steven10137,
can you please tell me the name of the book from where you read this formula
of gradient and correlation.
 

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