Transcript Ch 3.2 - Calculating LSRL

3.2 - Residuals and
Least Squares
Regression Line
Residuals
• the difference between an observed
value of the response variable and the
value predicted by the regression line.
• residual = y -ŷ
predicted long jump distance = 304.56 - 27.63(sprint time)
Find and interpret the residual for a sprint time of 8.09 seconds.
(FYI - the data gathered showed a long jump distance of 151 in for a sprint
time of 8.09 seconds)
First, find the predicted value
Second, find the residual.
Third, determine if it is above or below the predicted value.
Fourth, write an interpretive sentence!
Least Squares
Regression Line
• the line that makes the sum of the
square of the residuals as small as
possible.
Web Applet #1:
http://hspm.sph.sc.edu/courses/J716/demos/LeastSquares/LeastSqu
aresDemo.html
Web Applet #2: http://bcs.whfreeman.com/tps4e/#628644__666392__
Calculating the Least-Squares Regression Line
ŷ  a  bx
sy
br
sx
a  y  bx
* These are on your formula sheet, just with
different notation instead of a and b*
What does the slope of the least-squares
regression line tell us?
• a change in 1 standard deviation in x
corresponds to a change of r standard
deviations in y.
• There is a close relationship between
correlation and slope!
The number of miles (in thousands) for 11 used
Hondas has a mean of 50.5 and a standard
deviation of 19.3. The advertised prices had a
mean of $14,425 and a standard deviation of
$1899. The correlation for these variables is r = 0.874. Find the equation of the least-squares
regression line and explain what the slope
represents.
Slope:
sy
 1899 
b  r  .874 
 86

 19.3 
sx
Intercept: a  y  bx  14, 425  (86)(50)  18, 768
LSRL: ŷ  18768  86x
sx  19.3
rsy  1660
For each additional 19.3 thousand miles we expect
the cost to decrease $1660.