HAWKES LEARNING SYSTEMS
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Section 12.2
Linear Regression
Copyright © 2008 by Hawkes Learning
Systems/Quant Systems, Inc.
All rights reserved.
HAWKES LEARNING SYSTEMS
Regression, Inference, and Model Building
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12.2 Linear Regression
Definitions:
• If r is statistically significant, then a regression line
can be used to make predictions regarding the data.
• Regression Line – the line from which the average
variation from the data is the smallest. Also known
as the line of best fit. There is only one regression
line for each data set.
b0 = y-intercept
b1 = slope
Other commonly used regression line formulas:
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12.2 Linear Regression
Slope:
When calculating the slope, round your answers to three
decimal places.
y-Intercept:
When calculating the y-intercept, round your answers to three
decimal places.
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12.2 Linear Regression
TI-84 Plus Instructions:
1. Press STAT, then EDIT
2. Type the x-variable values into L1
3. Type the y-variable values into L2
4. Press STAT, then CALC
5. Choose 4: Linreg(ax+b)
6. Press ENTER
a = b1 = slope
b = b0 = y-intercept
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12.2 Linear Regression
Determine the regression line:
The local school board wants to evaluate the relationship between class size
and performance on the state achievement test. They decide to collect data
from various schools in the county. A portion of the data collected is listed
below. Each pair represents the class size and corresponding average score
on the achievement test for 8 schools.
Class Size
15
17
18
20
21
24
26
29
Average Score 85.3 86.2 85.0 82.7 81.9 78.8 75.3 72.1
Solution:
First decide which variable should be the x variable and which
variable should be y.
Class size is the independent variable, x, and average score is
the dependent variable, y.
We can use a calculator or the equations to find the regression
line.
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12.2 Linear Regression
Solution (continued):
Using the equations for slope and y-intercept:
n = 8, ∑x = 170, ∑y = 647.3, ∑xy = 13588.7, ∑x 2 = 3772
–1.043
103.085
Now place the slope and y-intercept into the regression line
equation:
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12.2 Linear Regression
Predictions:
A prediction should not be made with a regression model if:
1. The correlation is not statistically significant.
2. You are using a value outside of the range of the
sample data.
3. The population is different than that of the sample data.
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12.2 Linear Regression
Calculate the prediction:
Use the equation of the regression line,
to predict what the average achievement test score will be for
the following class size:
a. 16
86.397
b. 19
83.268
c. 25
77.010
d. 45
Since 45 is outside the range of the original data of
15 to 29, we cannot predict the test score.
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12.2 Linear Regression
Putting it all together:
The table below gives the average monthly temperature and corresponding
monthly precipitation totals for one year in Key West, Florida.
Temperatures (in °F) and Precipitation (in inches) in Key West, FL
Average
Temperature
Inches in
Precipitation
75
76
79
82
85
88
89
90
88
85
81
77
2.22 1.51 1.86 2.06 3.48 4.57 3.27 5.40 5.45 4.34 2.64 2.14
a. Create a scatter plot.
b. Calculate the correlation coefficient, r.
c. Verify that the correlation coefficient is statistically significant at the 0.05
level of significance.
d. Calculate the equation of the line of best fit.
e. Calculate and interpret the coefficient of determination, r 2.
f. If appropriate, make a prediction for the monthly precipitation for a
month in which the average temperature is 80 degrees.
g. If appropriate, make a prediction for the monthly precipitation in Destin,
FL for a month in which the average temperature is 83 degrees.
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12.2 Linear Regression
Solution (continued):
a. Create a scatter plot.
The scatter plot shows a positive trend in data.
HAWKES LEARNING SYSTEMS
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12.2 Linear Regression
Solution (continued):
b. Calculate the correlation coefficient, r.
n = 12, ∑x = 995, ∑y = 38.94, ∑xy = 3299.23, ∑x2 = 82815,
∑y2 = 147.847
 0.859
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12.2 Linear Regression
Solution (continued):
c. Verify that the correlation coefficient is statistically
significant at the 0.05 level of significance.
n = 12, r  0.859, a = 0.05
ra = 0.576
Statistically significant if |r | > ra .
Therefore, r is statistically significant at the 0.05 level.
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12.2 Linear Regression
Solution (continued):
d. Calculate the equation of the line of best fit.
n = 12, ∑x = 995, ∑y = 38.94, ∑xy = 3299.23, ∑x2 = 82815
0.225
–15.424
Now place the slope and y-intercept into the regression line
equation:
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12.2 Linear Regression
Solution (continued):
e. Calculate and interpret the coefficient of determination, r 2.
r  0.859
r 2  0.738
So about 73.8% of the variation in precipitation can be
attributed to the linear relationship between temperature and
precipitation. The remaining 26.2% of the variation is from
unknown sources.
HAWKES LEARNING SYSTEMS
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12.2 Linear Regression
Solution (continued):
f. If appropriate, make a prediction for the monthly
precipitation for a month in which the average temperature
is 80 degrees.
2.576
Thus a reasonable estimate for the precipitation for a month in
which the average temperature is 80 degrees is about 2.58
inches.
g. If appropriate, make a prediction for the monthly
precipitation in Destin, FL for a month in which the average
temperature is 83 degrees.
The data was collected in Key West, not Destin, FL.
Therefore, it is not appropriate to use the linear regression line
to make predictions regarding the precipitation in Destin.