linear regression larson/farber 4th ed. 1 section 9.2

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Linear Regression Larson/Farber 4th ed. 1 Section 9.2

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Page 1: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

Linear Regression

Larson/Farber 4th ed.1

Section 9.2

Page 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

Page 3: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

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

Page 4: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

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

Page 5: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

Regression Line

Larson/Farber 4th ed.5

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

Page 6: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

The Equation of a Regression Line

Larson/Farber 4th ed.6

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

Page 7: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

Example: Finding the Equation of a Regression Line

Larson/Farber 4th ed.7

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

Page 8: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

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:

Page 9: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

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

Page 10: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

Solution: Finding the Equation of a Regression Line

Larson/Farber 4th ed.10

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

Page 11: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

Example: Predicting y-Values Using Regression Equations

Larson/Farber 4th ed.11

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

Page 12: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

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

Page 13: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

Solution: Predicting y-Values Using Regression Equations

Larson/Farber 4th ed.13

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).

Page 14: Linear Regression Larson/Farber 4th ed. 1 Section 9.2

Section 9.2 Summary

Larson/Farber 4th ed.14

Found the equation of a regression linePredicted y-values using a regression

equation