Download - Section 9.2
Linear Regression
Larson/Farber 4th ed.1
Section 9.2
Section 9.2 Objectives
Larson/Farber 4th ed.2
Find the equation of a regression linePredict y-values using a regression
equation
Regression lines
Larson/Farber 4th ed.3
After verifying that the linear correlation between two variables is significant, next we determine the equation of the line that best models the data (regression line).
Can be used to predict the value of y for a given value of x.
x
y
Residuals
Larson/Farber 4th ed.4
ResidualThe difference between the observed y-value
and the predicted y-value for a given x-value on the line.
For a given x-value,
di = (observed y-value) – (predicted y-value)
x
y
}d1
}d
2
d3
{
d4{ }d
5
d6{
Predicted y-value
Observed y-value
Regression Line
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Regression line (line of best fit)The line for which the sum of the squares of
the residuals is a minimum. The equation of a regression line for an
independent variable x and a dependent variable y is
ŷ = mx + b
Predicted y-value for a given x-value
Slope
y-intercept
The Equation of a Regression Line
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ŷ = mx + b where
is the mean of the y-values in the data is the mean of the x-values in the dataThe regression line always passes through
the point
22
n xy x ym
n x x
y xb y mx mn n
y
x
,x y
Example: Finding the Equation of a Regression Line
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Find the equation of the regression line for the advertising expenditures and company sales data.
Advertisingexpenses,($1000), x
Companysales
($1000), y2.4 2251.6 1842.0 2202.6 2401.4 1801.6 1842.0 1862.2 215
Solution: Finding the Equation of a Regression Line
Larson/Farber 4th ed.8
x y xy x2 y2
2.4 2251.6 1842.0 2202.6 2401.4 1801.6 1842.0 1862.2 215
540294.4440624252
294.4372473
5.762.56
46.761.962.56
44.84
50,62533,85648,40057,60032,40033,85634,59646,225
Σx = 15.8 Σy = 1634 Σxy = 3289.8 Σx2 = 32.44 Σy2 = 337,558
Recall from section 9.1:
Solution: Finding the Equation of a Regression Line
Larson/Farber 4th ed.9
Σx = 15.8 Σy = 1634 Σxy = 3289.8 Σx2 = 32.44 Σy2 = 337,558
22
n xy x ym
n x x
b y mx
28(3289.8) (15.8)(1634)
8(32.44) 15.8
501.2 50.728749.88
1634 15.8(50.72874)8 8
204.25 (50.72874)(1.975) 104.0607
ˆ 50.729 104.061y x Equation of the regression line
Solution: Finding the Equation of a Regression Line
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To sketch the regression line, use any two x-values within the range of the data and calculate the corresponding y-values from the regression line.
ˆ 50.729 104.061y x
x
Advertising expenses
(in thousands of dollars)
Com
pan
y s
ale
s(i
n t
hou
san
ds
of
dolla
rs)
y
Example: Predicting y-Values Using Regression Equations
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The regression equation for the advertising expenses (in thousands of dollars) and company sales (in thousands of dollars) data is ŷ = 50.729x + 104.061. Use this equation to predict the expected company sales for the following advertising expenses. (Recall from section 9.1 that x and y have a significant linear correlation.)1.1.5 thousand dollars2.1.8 thousand dollars3.2.5 thousand dollars
Solution: Predicting y-Values Using Regression Equations
Larson/Farber 4th ed.12
ŷ = 50.729x + 104.0611. 1.5 thousand dollars
When the advertising expenses are $1500, the company sales are about $180,155.
ŷ =50.729(1.5) + 104.061 ≈ 180.155
2. 1.8 thousand dollars
When the advertising expenses are $1800, the company sales are about $195,373.
ŷ =50.729(1.8) + 104.061 ≈ 195.373
Solution: Predicting y-Values Using Regression Equations
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3. 2.5 thousand dollars
When the advertising expenses are $2500, the company sales are about $230,884.
ŷ =50.729(2.5) + 104.061 ≈ 230.884
Prediction values are meaningful only for x-values in (or close to) the range of the data. The x-values in the original data set range from 1.4 to 2.6. So, it wouldnot be appropriate to use the regression line to predictcompany sales for advertising expenditures such as 0.5 ($500) or 5.0 ($5000).
Section 9.2 Summary
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Found the equation of a regression linePredicted y-values using a regression
equation