The line of best fit.

Step 1 of 5

By hand

A correlation says how strong a link is. Sometimes you want more: a line that predicts $y$⁠ for any $x$⁠. In the linear functions Unit you met the line $y = a + bx$⁠, with slope $b$⁠ and intercept $a$⁠. Which such line fits the points best?

Every line misses every point by some vertical distance, its residual. On the board each residual is a stick from the bead to the line, with a square on each stick. The best line is the one whose squares add up to the least: the line of best fit, by least squares. Squares and not plain distances, for the same reason as in the standard deviation: distances above and below would cancel.

Your line turns about $(\bar{x}, \bar{y})$⁠, and the coloured square under the plot is the average square about it, $s_e^2$⁠. The chart beside the Scene shows the squares’ area for every slope. Find the best line.

Step 2 of 5

The balance point

The board now turns your line to the exact line of best fit and prints it: $\hat{y} = 1.90 + 0.62x$$\hat{y} =$$1.90 + 0.62x$⁠. The hat on $\hat{y}$⁠ means “predicted value”, as against a value that was measured.

The slope of the line of best fit is

$b = \frac{\sum (x - \bar{x})(y - \bar{y})}{\sum (x - \bar{x})^2}$$b =$$\frac{\sum (x - \bar{x})(y - \bar{y})}{\sum (x - \bar{x})^2}$⁠

But first, a question about the point the line turned about.

Step 3 of 5

Predicting

This is what the line is for: predicting $y$⁠ at an $x$⁠ that was not measured exactly. Put the $x$⁠ into the equation.

A prediction inside the range of the readings, here between $1$⁠ and $9$⁠, is called interpolation, and it is reasonable. A prediction far outside it, extrapolation, is risky: no reading tells you the link stays a straight line out there.

Step 4 of 5

One far point

In the Lesson on averages you saw that an outlier pulls the mean and leaves the median nearly where it was. The line of best fit is built from squares, so it behaves like the mean, not the median.

Step 5 of 5

How much the line explains

The board also prints $r^2 = 1 - \frac{s_e^2}{s^2}$$r^2 =$$1 - \frac{s_e^2}{s^2}$⁠. The two squares under the plot compare the average square about the mean, $s^2$⁠, with the average square about the line, $s_e^2$⁠. $r^2$⁠ is the share of the scatter in $y$⁠ the line accounts for, and $1 - r^2$⁠ the share left over as noise. Here $r^2 = 0.78$⁠: the line accounts for $78\%$⁠ of the scatter.

Move Noise and watch the two squares close up or draw apart. To finish, work out $r^2$⁠ for a real study.