1. Scatter Plots
&
Lines of Fit
Algebra Glencoe McGraw-Hill JoAnn Evans

2. One way to accomplish this is to use a graphing calculator.
You have studied how to find an equation of a line that passes through two points. For each of those problems, there was only one line that could pass through the two points in question.
In this lesson there will be MULTIPLE points and no one line will pass through all of them. Your job will be to find the line that most closely FITS the points.
One way to accomplish this is to use a graphing calculator.
Another way is to use a computer and statistical formulas to find the best fitting line.
However, we will be using a graphical approach to approximate a line of fit…

3. Plot the following points on a coordinate plane:
(10, 12)
(16, 31)
(13, 4)
(-18, -37)
(14, 26)
(-1, 0)
(2, -1)
(-14, -32)
(3, 2)
(6, 8)
(-5, -20)
(-8, -23)
In order to have enough room for all the points on the graph, number your axes in multiples of 5.

4. Graph the line y = 2x – 5 on the same coordinate plane.
Try moving your ruler around until you find a line you think might be a line of fit for the data points. 10
How about this one? -5 5
Graph the line y = 2x – 5 on the same coordinate plane.
-10
It looks like a good line of fit.
How many points lie above the line? Below? On?

5. To Find a Line of Fit:
1. Carefully plot the data points for a situation.
2. Use a ruler to sketch the line you think best approximates the data points. Try to have some points above, some below, and maybe a few on the line.
3. Identify two points that lie on the line (they don’t have to be two of the original data points). Approximate the x- and y-coordinates for both those points.
4. Find the slope between those two points, then the equation of the line that connects the two points.

6. The winning Olympic times for the 100-meter run from 1928 to 1988 are shown in this table.
Approximate a line of fit for these times. Let y represent the winning time and x represent the year (x = 0 will correspond to 1928).
Year
Time in seconds
Point to graph
1928
12.2
(0, 12.2)
1932
11.9
(4, 11.9)
1936
11.5
(8, 11.5)
1948
(20, 11.9)
1952
(24, 11.5)
1956
(28, 11.5)
1960
11.0
(32, 11)
1964
11.4
(36, 11.4)
1968
(40, 11)
1972
11.07
(44, )
1976
11.08
(48, )
1980
11.6
(52, 11.6)
1984
10.97
(56, )
1988
10.54
(60, )

7. winning time # of Years After 1928 12.2 12.0 11.8 11.6 11.4 11.2 11.0
10.8
10.6
# of Years After 1928

8. One possible line of fit would contain the points (0, 12) and (28, 11
Calculate the slope between these points.
The y-intercept is 12.0, so an equation of the line of fit could be:

9. Determining the CORRELATION of x and y:
x and y have a positive correlation—the line of fit will have a positive slope x
x and y have a negative correlation—the line of fit will have a negative slope y
x and y have NO correlation—and there is NO line of fit x

10. The tables contain the forearm lengths and foot lengths (without shoes) of 21 students in an algebra class. Plot the points on the graph provided.
Forearm
Foot
22 cm
24 cm
20 cm
19 cm
21 cm
23 cm
25 cm
18 cm
22 cm
19 cm
25 cm
23 cm
24 cm
20 cm
21 cm
18 cm
27 cm

11. Comparing Forearm Length to Foot Length27 26 25 24
Foot length in cm. 23 22 21 20 19 18
Forearm length in cm.

12. Don’t sketch the line until after our discussion.
Using the graph provided and a ruler, experiment to find a line of fit. Stand your ruler on its side and look down at the paper. Move the ruler around until you have a line that seems to fit well. Count the number of points above, below, and on the line.
Don’t sketch the line until after our discussion.

13. Finding a Line of Fit Foot length in cm. Forearm length in cm. 27 2625 24
Foot length in cm. 23 22 21 20 19 18
Forearm length in cm.

14. Find the equation of the line that contains the points (19, 20) and (26, 26).
Find the slope: (26, 26)
(19, 20)
Substitute the values m = .86, x = 19, and y = 20 into the slope-intercept form.