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# Scatter Plots

## Key Concepts

After this lesson student will be able to:

• Understand association.
• Understand correlation.
• Write the equation of a trend line.
• Interpret trend lines.

### Scatter Plots

#### Model and discuss

Nicholas plotted data points to represent the relationship between screen size and the cost of television sets. Everything about televisions is the same except for the screen size.

1. Describe any patterns you see.
2. What does this set of points talk about the relationship of screen size and the cost of television?
3. Where do you think the point for a 46-inch television would be on the graph? How about a 60-inch TV? Explain?

Solution:

1. As the screen size of the television set increases, the cost also increases.
2. The relationship of the screen size and the cost of the television is approximately linear.
3. The cost of a 46-inch TV should be somewhere in between the cost of the 43- and 48-inch TV.

So, we can approximate its cost at about \$450.

The cost of a 60-inch TV should be greater than that of a 50-inch TV.

So, we approximate the cost of a 60-inch TV to be approximately more than \$700.

### Understand Association

Example 1.

1. What is the relationship between the hours after sunrise, x, and the temperature, y, shown in the scatter plot?

Solution:

As the number of hours after sunrise increases, so does the temperature.

When y-values tend to increase as x-values increase, the two data sets have a positive association.

1. What is the relationship between the hours after sunset, x, and the temperature, y, shown in the scatter plot?

Solution:

As the number of hours after sunset increases, the temperature decreases.

When y-values tend to decrease as x-values increase, the two data sets have a negative association.

1. What is the relationship between the hours after sunset, x, and the amount of rain, y, shown in the scatter plot?

Solution:

There is no relationship between the amount of rainfall and the number of hours after sunset.

When there is no general relationship between x-values and y-values, the two data sets have no association.

#### Try it

1. Describe the type of association each scatter plot shows.

Solution:

If the y-values increase as the x-values increase, then there is a positive association

If the y-values decrease as the x-values increase, then there is a negative association.

If there is no relationship between the x- and y-values, then there is no association.

1. As the x-values increase, the y-values also increase, so there is a positive association.
1. There is no relationship between the x- and y-values, so there is no association.

### Understand correlation

Example 2

1. How can the relationship between the hours after sunrise, x, and the temperature, y, be modeled?

Solution:

The data points on the scatter plot approximates a line. Sketch a trend line that best fits the data to determine whether a linear function can model the relationship.

The scatter plot suggests a linear relationship.

There is a positive correlation between hours after sunrise and the temperature.

When data with a negative association are modeled with a line, there is a negative correlation.

If the data do not have an association, they cannot be modeled with a linear function.

#### Try it

1. How can the relationship between the hours after sunset, x, and the temperature, y, be modeled?

If the relationship is modeled with a linear function, describe the correlation between the two data sets.

Solution:

The data points on the scatter plot approximate a line, so, it suggests a linear relationship. As the number of hours after sunset increases, the temperature decreases so there is a negative correlation between hours after sunset and the temperature.

Linear, negative correlation

### Write the equation of a trend line

Example 3

What trend line models the data in the scatter plot?

Solution:

A trend line models the data in a scatter plot by showing the general direction of the data. A trend line fits the data as closely as possible.

Step 1:

Sketch a trend line for the data.

A trend line approximates a balance of

points above and below that line.

Step 2:

Write the equation of this trend line.

Select two points on the trend line to find the slope.

Use the slope and one of the points to write the equation in slope intercept form.

y – 50 = 32.1(x-2)

y = 32.1x – 14.2

This trend line that models the data is y = 32.1x – 14.2

This trend line is one of many possible trend lines.

#### Try it

3.

1. What trend line, in slope-intercept form, models the data from example 2 try it?
2. Explain why there could be no points on a trend line, yet the line models the data?

Solution:

1. Depending on the trend line drawn, a possible two points to be used are (0, 71) and (4, 66).

The slope of the trend line is:

m= 66−71 / 4−0

m = −5 / 4

m=-1.25

Use the slope and one of the points to write the equation in slope-intercept form:

y — y₁ = m (x − x₁)

y – 71 = -1.25(x – 0)

y = -1.25x + 71

1. Because a trend line approximates a balance of points above and below the line.

### Interpret trend lines

Example 4.

The table shows the amount of time required to download a 100-megabyte file for various internet speeds. Assuming the trend continues, how long would it take to download the 100-megabyte file if the internet speed is 75 kilobytes per second?

Solution:

Step 1

Make a scatter plot of the data and sketch a trend line.

Step 2

Find the equation of the trend line.

Select the two points on the trend line to find the slope:

(40, 6) and (55, 4)

= 6−4 / 40−55 ≈ -0.13

Use the slope and one of the points to write the equation for the trend line.

y – 6 = -0.13(x – 40)

y = -0.13x + 11.2

The equation of the trend line is y = -0.13x + 11.2.

Step 3

Use the equation of the linear model to find the y-value that corresponds to x = 75.

y = -0.13x + 11.2

y = -0.13(75) + 11.2

y = 1.45

#### Try it

1. What is the x-intercept of the trend line? Is that possible in a real–world situation? Explain.

Solution:

From example 4, we have the equation: y = – 0.13x + 11.2

To find the x-intercept, we set y = 0 and solve for x:

0 = – 0.13x + 11.2

0.13x = 11.2

x ≈ 86

This represents an internet speed of 86 KB/s, which is a possible internet speed but since the corresponding time is 0 minutes, then this is not possible in a real-world situation.

• The table shows the maximum recommended viewing distances y, in feet, for an HDTV with screen size x, in inches. What trend line models the data shown in the table? What does the slope of the trend line represent?

Solution:

Make a scatter plot of the data and sketch a trend line.

Find the equation of the trend line.

Select two points on the trend line to find the slope.

Here, let’s use (60, 12.5) and (40, 12.5):

m= 12.5−8.3 / 60−40

m=0.21

Use the slope and one of the points to write the equation in slope-intercept form:

y – y₁ = m (x − x₁)

y – 8.3 = 0.21(x – 40)

y – 8.3 = 0.21x – 8.4

y = 0.21x – 0.1

The slope means that the maximum viewing distance increases by about 0.21 ft as the screen size of an HDTV increases by 1 inch.

1. Describe the type of association between x and y for each set of data. Explain.
1. Describe the type of association between x and y for the set of data. Explain.
1. How could be finding the y-intercept of a trend line for a data set help you determine the usefulness of the trend line as a model?
2. Describe the type of association each scatter plot shows.
1. Make a scatter plot of the data. Describe the type of association that the scatter plot shows.
1. Make a scatter plot of the data. Draw a trend line and write its equation.

1. If you observe, the graph y-values increase as x -value increases. So, the graph has a positive association
1. As the x-values increase, the y-values decrease, so, there is a negative association.
1. The y-intercept gives us the initial value of the situation; so by using it, we can determine if the trend line is a good fit for the data set.
1. If you observe the graph, the y-values increase and decrease as x-value increases. So, it has no association.
1. Scatter plot of the given data is from the data, y-values decrease as x-value increase, then it has a negative association.
1. Find the equation of the trend line. Select the two points on the trend line to find the slope. Here, I decided to use (4, 5) and (10, 10):

m= 10−5 / 10−4

m= 5 / 6

Use the slope and one of the points to write the equation in slope-intercept form:

y-y₁ = m (x – x₁)

y-5= 5 / 6(x-4)

y – 5 = 5x / 6 – 5X4 / 6

y – 5 = 5x / 6 – 10 / 3

y =5x / 6 +  5 / 3

## Exercise

• Describe the type of association between x and y for each set of data. Explain
• Tell whether the data show a positive, a negative, or no relationship. Television size and price
• Make a scatter plot of the data. Tell whether the data shows a positive, a negative or, no relationship.

### Concept Summary

#### Scatter Plots and Trend Lines

Positive Association

Negative Association

### Concept Map

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