least squares regression line

y=mx+b

y=a+bx

$y^{\wedge}=a+bx$

plug in a predicted guess for y-hat and simplify

look at the scatterplot

rearrange the equation to solve for x

plug in a value for x and simplify

the difference between the actual value of y and the value of y predicted by the regression line

actual y - predicted y

$y-y^{\wedge}$

the difference between the actual value of x and the value of x predicted by the regression line

When asked to interpret the slope or y intercept, it is very important to include the word predicted (or equivalent) in your response.

a is the y-intercept

using the regression line to make a prediction for x=0 is an extrapolation

b is the slope, the amount by which the predicted value of y

changes when x increases by 1 unit

good

bad

ugly

mean

residual

predicted

estimated

left overs

If there is no leftover curved pattern in the residual plot, the regression model is appropriate.

If there is leftover curved pattern in the residual plot, the regression model is appropriate.

If there is a leftover curved pattern in the residual plot, consider using a regression model with a different form.

If there is not a leftover curved pattern in the residual plot, consider using a regression model with a different form.

The standard deviation of the residuals s measures the size of a typical residual.

$s=\sqrt{\frac{sum\ of\ squared\ residuals}{n-2}}$

$s=\sqrt{\frac{\sum_{ }^{ }\left(y_i-y_i^{\wedge}\right)^2}{n-2}}$

That is, s measures the typical distance between the actual y values and the predicted y values.