Association, Clusters, and Outliers in Scatter Plots | Grade 8 Math | LevelKite

You already know how to read a point $(x, y)$ on a coordinate plane. A scatter plot uses points like these to show data that come in pairs. For each person or object, you record two quantities, and each pair of values is shown as one point.

Here is your first data set. Eight basketball players each recorded two numbers. Let $x$ be the time, in hours, that a player practiced free throws last week, so $x \ge 0$ . Let $y$ be the number of free throws the player then made in a test of 10 shots, so $y$ is a whole number from 0 to 10. The horizontal axis shows hours of practice, and the vertical axis shows free throws made.

The players give the points $(1, 2)$ , $(2, 4)$ , $(3, 3)$ , $(4, 5)$ , $(5, 6)$ , $(6, 5)$ , $(7, 8)$ and $(8, 9)$ . For example, $(1, 2)$ stands for a player who practiced for 1 hour and made 2 of the 10 free throws.

Each point is one player: hours of practice (

x

) and free throws made out of 10 (

y

), from

(1, 2)

to

(8, 9)

.

The association between two quantities describes how one tends to change as the other increases. To see it, read the points from left to right, in order of increasing $x$ , and ask what the $y$ -values do overall.

Positive association: as $x$ increases, $y$ tends to increase, so the points tend to rise from left to right.
Negative association: as $x$ increases, $y$ tends to decrease, so the points tend to fall from left to right.
No association: the points show no overall rise or fall. High and low $y$ -values are mixed across all the $x$ -values, so knowing $x$ does not help you predict $y$ .

The words tends to matter. An association describes the overall pattern, the general way most of the points go together. That description is approximate, so some pairs of points can go against it. An association also tells you only how the values go together. It does not prove that a change in one quantity causes a change in the other.

Worked example

Practice time and free throws

Describe the association between hours of practice ( $x$ ) and free throws made ( $y$ ) for the eight players above.

List the free-throw counts in order of increasing practice time, because association is about what one quantity does as the other increases.
$2 \to 4 \to 3 \to 5 \to 6 \to 5 \to 8 \to 9$
Compare the four players who practiced 1 to 4 hours with the four who practiced 5 to 8 hours. Find the mean of each group, because a mean gives one typical value for a group.
$\frac{2+4+3+5}{4} = \frac{14}{4} = 3.5 \qquad \frac{6+5+8+9}{4} = \frac{28}{4} = 7$
The players who practiced more made more free throws on average, 7 compared with 3.5. The counts dip twice, from 4 to 3 and from 6 to 5, but an association describes the overall pattern, not every pair of points. So free throws made tend to increase as practice time increases.
$7 > 3.5$
Interpret with care. The plot alone does not prove that practicing causes a player to make more free throws. For example, players who already shoot well might simply choose to practice more.

Answer

There is a positive association: players who practiced more tended to make more free throws, although the plot does not prove that the practice caused this.

Next, pair the same eight free-throw counts with a different quantity. Let $x$ be the number of letters in a player's first name, a whole number shown on the horizontal axis. As before, $y$ is the number of free throws made out of 10, shown on the vertical axis. This time the players give the points $(3, 4)$ , $(4, 9)$ , $(5, 2)$ , $(6, 6)$ , $(7, 3)$ , $(8, 8)$ , $(9, 5)$ and $(10, 5)$ .

Each point is one player: letters in first name (

x

) and free throws made out of 10 (

y

), from

(3, 4)

to

(10, 5)

.

Worked example

Name length and free throws

Describe the association between name length ( $x$ ) and free throws made ( $y$ ) for the eight points above.

List the free-throw counts in order of increasing name length. They jump up and down with no steady direction.
$4 \to 9 \to 2 \to 6 \to 3 \to 8 \to 5 \to 5$
Split the players into the four with the shortest names (3 to 6 letters) and the four with the longest names (7 to 10 letters), and find the mean of each group.
$\frac{4+9+2+6}{4} = \frac{21}{4} = 5.25 \qquad \frac{3+8+5+5}{4} = \frac{21}{4} = 5.25$
Neither group tends to make more free throws: the means match here, and each group mixes low and high counts, 2 and 9 for the shorter names and 3 and 8 for the longer ones. So free throws made show no tendency to rise or fall as name length increases.

Answer

There is no association: name length does not help you predict free throws made. The counts are the same eight numbers as before, so the association depends on which quantity they are paired with.

Key concept

Describing a scatter plot

Describe a scatter plot by its overall pattern, the general way most of the points go together. The pattern is approximate, so it does not have to fit every point.

State the association: positive, negative or none. It describes a tendency, not proof of cause.
Look for clusters: groups of points that lie close together, separated from other points by an empty stretch called a gap.
Look for outliers: points that lie far from the overall pattern of the rest of the data.

Support each statement with coordinates, not just with how the plot looks.

Now use all three parts. A used-car lot lists 11 cars. Let $x$ be a car's age in years, shown on the horizontal axis, and let $y$ be its value in thousands of dollars, shown on the vertical axis. Both are at least 0, and $y = 10$ means 10,000 dollars. The cars give the points $(1, 11)$ , $(1, 10)$ , $(2, 10)$ , $(2, 9)$ , $(3, 8)$ , $(7, 5)$ , $(7, 4)$ , $(8, 4)$ , $(9, 3)$ , $(10, 2)$ and $(9, 10)$ . In the plot below, Point A is $(9, 10)$ and Point B is $(10, 2)$ .

Each point is one car: age in years (

x

) and value in thousands of dollars (

y

), with Point A at

(9, 10)

and Point B at

(10, 2)

.

Worked example

Ages and values of used cars

Describe the scatter plot of the 11 cars above. State the association, identify any clusters and identify any outliers, using coordinates as evidence.

Find the direction. The five cars aged 1 to 3 years are worth 8 to 11 thousand dollars, while five of the six cars aged 7 to 10 years are worth only 2 to 5 thousand. Value tends to decrease as age increases, which is a negative association. Like any association, it does not by itself prove that age causes the lower values.
Look for gaps. No car is 4, 5 or 6 years old, so the points form two clusters with a gap between them: the five cars aged 1 to 3 in the upper left, and, in the lower right, the five cars aged 7 to 10 that are worth 2 to 5 thousand dollars, including Point B. So the lot has a group of newer cars and a group of older cars, with none in between.
$\text{no points with } 3 < x < 7$
Check Point A, at (9, 10), against the pattern. The other 9-year-old car, at (9, 3), is worth 3 thousand dollars, so Point A is worth 7 thousand dollars more than a car of the same age.
$10 - 3 = 7$
Check whether Point A fits either cluster. The upper-left cluster's oldest car is 3 years old, and the lower-right cluster's most valuable car is worth 5 thousand dollars. Point A is at least 6 years older than every car in the first cluster and worth at least 5 thousand more than every car in the second. It lies far from the pattern of the rest, so it is an outlier.
$9 - 3 = 6 \qquad 10 - 5 = 5$

Answer

The plot shows a negative association between age and value, two clusters separated by a gap where no car is 4 to 6 years old, and one outlier, Point A at $(9, 10)$ .

To describe a scatter plot, first state its association: positive, negative or none. The association is a tendency of the overall pattern, not a rule for every point, and it is not proof of cause. Then name any clusters and gaps, and any outliers far from the pattern, and back up each claim with coordinates. Next, you will construct scatter plots from data yourself.