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lexber
 one year ago
how do i make a line of best fit?
lexber
 one year ago
how do i make a line of best fit?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use the method of least squares to find a regression line: https://en.wikipedia.org/wiki/Least_squares

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Of course it depends on what kind of "best fit" you're looking for. Do you have a data set you're trying to model with linear function?

lexber
 one year ago
Best ResponseYou've already chosen the best response.0i need to make a line of best fit out of this and write the approximate slope but im not understanding how to do it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You're looking for a line of the form \(\hat{y}=\beta_0+\beta_1x\), and \(\beta_1\) is the slope you want in the end. I don't know where you're at in terms of the theory behind regression, so for the sake brevity, you can determine the slope of the regression line using this formula: \[\beta_1=\frac{\displaystyle\sum_{i=1}^n (x_i\bar{x})(y_i\bar{y})}{\displaystyle\sum_{i=1}^n(x_i\bar{x})^2}\]where \(\bar{x}=\displaystyle\frac{1}{n}\sum_{i=1}^nx_i\) (the average of all the xcoordinates of the given points). The same goes for \(\bar{y}\), just replace \(x_i\) with \(y_i\). The plot gives you the points \((x_i,y_i)\) you need to compute the averages. So for example, the three bottomleftmost points are \((2.5,1)\), \((5,2)\), and \((7.5,2)\). Your averages would be \[\bar{x}=\frac{2.5+5+7.5}{3}=5\quad\quad\quad\bar{y}=\frac{1+2+2}{3}\approx1.67\] Then the best fit line FOR THESE THREE POINTS ONLY would be \[\begin{align*}\beta_1&=\frac{(2.55)(11.67)+(55)(21.67)+(7.55)(21.67)}{(2.55)^2+(55)^2+(7.55)^2}\\[1ex] &=0.2\end{align*}\] You have 17 points to account for, but this is basically a template for what you have to do.
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