Simple Linear Regression 1. the least squares procedure 2. inference for least squares lines

1

Simple Linear Regression1. the least squares procedure

2. inference for least squares lines

Chapter 12

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Introduction

• We will examine the relationship between quantitative variables x and y via a mathematical equation.

• The motivation for using the technique:– Forecast the value of a resposne variable (y) from

the value of explanatory variables (x1, x2,…xk.).– Analyze the specific relationship between the

explanatory variable and the dependent variable.

3

House size

HouseCost

Most lots sell for $25,000

Building a house costs about

$75 per square foot.

House cost = 25000 + 75(Size)

The Model

The model has a deterministic and a probabilistic component

4

House cost = 25000 + 75(Size)

House size

HouseCost

Most lots sell for $25,000

+ e

However, house costs vary even among same size houses!

The Model

Since cost behave unpredictably, we add a random component.

5

The Model

• The first order linear model

y = response variablex = explanatory variableb0 = y-interceptb1 = slope of the linee = error variable

e+b+b xy 10

x

y

b0Run

Rise b1 = Rise/Run

b0 and b1 are unknown populationparameters, therefore are estimated from the data.

6

Estimating the Coefficients

• The estimates are determined by – drawing a sample from the population of interest,– calculating sample statistics.– producing a straight line that cuts into the data.

wwww w w w w

w

w ww

w ww

Question: What should be considered a good line?

x

y

7

The Least Squares (Regression) Line

A good line is one that minimizes the sum of squared differences between the points and the line.

8

The Least Squares (Regression) Line

3

3

ww

ww

41

1

4

(1,2)

2

2

(2,4)

(3,1.5)

Sum of squared differences = (2 - 1)2 + (4 - 2)2 +(1.5 - 3)2 +

(4,3.2)

(3.2 - 4)2 = 6.89Sum of squared differences = (2 -2.5)2 + (4 - 2.5)2 + (1.5 - 2.5)2 + (3.2 - 2.5)2 = 3.99

2.5

Let us compare two linesThe second line is horizontal

The smaller the sum of squared differencesthe better the fit of the line to the data.

9

The Estimated CoefficientsTo calculate the estimates of the slope and intercept of the least squares line , use the formulas:

1

0 1

2

1

2

1

correlation coefficient

( )

1

( )

1

y

x

n

ii

y

n

ii

x

sb r

sb y b xr

y ys

n

x xs

n

The least squares prediction equation that estimates the mean value of y for a particular value of x is:

0 1y b b x +

10

• Example:– A car dealer wants to find

the relationship between the odometer reading and the selling price of used cars.

– A random sample of 100 cars is selected, and the data recorded.

– Find the regression line.

Car Odometer Price1 37400 146002 44800 141003 45800 140004 30900 156005 31700 156006 34000 14700

. . .

. . .

. . .

Independent variable x

Dependent variable y

The Simple Linear Regression Line

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The Simple Linear Regression Line• Solution

– Solving by hand: Calculate a number of statistics36,011;

14,841;

x

y

6596.125

547.74

x

y

s

s

where n = 100.

1

0 1

547.740.81517 .067696596.125

14,841 ( .06769)(36,011) 17, 286.15

y

x

sb r

sb y b x

0 1ˆ 17, 286.15 .06769y b b x x +

0.80517r

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• Solution – continued– Using the computer

1. Scatterplot2. Trend function3. Data tab > Data Analysis > Regression


Regression StatisticsMultiple R 0.805167979R Square 0.648295475

Adjusted R Square 0.644706653Standard Error 326.4886258Observations 100

ANOVA df SS MS F Significance F

Regression 1 19255607.37 19255607.37 180.643 5.75078E-24Residual 98 10446292.63 106594.8228Total 99 29701900

Coefficients Standard Error t Stat P-value Lower 95% Upper 95%Intercept 17248.72734 182.0925742 94.72504534 3.57E-98 16887.37056 17610.084Odometer -0.06686089 0.004974639 -13.44034928 5.75E-24 -0.076732895 -0.0569889

13

ˆ 17, 248.73 .06686y x


ˆ 17, 248.73 .06686y x

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This is the slope of the line.For each additional mile on the odometer,the price decreases by an average of $0.0669

Odometer Line Fit Plot

13000

14000

15000

16000

Odometer

Pric

e

Interpreting the Linear Regression -Equation

The intercept is b0 = $17248.73.

0 No data

Do not interpret the intercept as the “Price of cars that have not been driven”

17248.73

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Error Variable: Required Conditions

• The error e is a critical part of the regression model.• Four requirements involving the distribution of e must

be satisfied.– The probability distribution of e is normal.– The mean of e is zero: E(e) = 0.– The standard deviation of e is se for all values of x.– The set of errors associated with different values of y are

all independent.

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The Normality of e

From the first three assumptions we have:y is normally distributed with meanE(y) = b0 + b1x, and a constant standard deviation se

m3

b0 + b1x1

b0 + b1x2

b0 + b1x3

E(y|x2)

E(y|x3)

x1 x2 x3

m1

E(y|x1)

m2

The standard deviation remains constant,

but the mean value changes with x

17

Assessing the Model

• The least squares method will produces a regression line whether or not there is a linear relationship between x and y.

• Consequently, it is important to assess how well the linear model fits the data.

• Several methods are used to assess the model. All are based on the sum of squares for errors, SSE.

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– This is the sum of differences between the points and the regression line.

– It can serve as a measure of how well the line fits the data. SSE is defined by

.)yy(SSEn

1i

2ii

Sum of Squares for Errors

20 1i i i iSSE y b y b x y

– A shortcut formula

19

– The mean error is equal to zero.– If se is small the errors tend to be close to zero

(close to the mean error). Then, the model fits the data well.

– Therefore, we can, use se as a measure of the suitability of using a linear model.

– An estimator of se is given by se

2

tan

nSSEs

EstimateofErrordardS

e

Standard Error of Estimate

20

• Example:– Calculate the standard error of estimate for the previous

example and describe what it tells you about the model fit.• Solution

10,446,293

10,446,293 326.492 98

SSE

SSEsne

It is hard to assess the model based on se even when compared with the mean value of y.

326.49 14,841s ye

Standard Error of Estimate,Example

21

q

qq

q

qq

qq

q

q

qq

qq

q

qq qq

q

q

q

Testing the slope– When no linear relationship exists between two

variables, the regression line should be horizontal.

q

q

q

qq

qqq

qq

q

q

qq

qq

qqq

qq

q

q

qq

qq

qqq

qq

q

q

qq

qq

qqq

qq

q

q

qq

qq

q

qq q

q

q

q

q

q

qq

q

qq qq

q

q

q

q

qq

q

qq q

q

q

q

q

q

qq

qqq

qq

q

q

q

Different inputs (x) yielddifferent outputs (y).

No linear relationship.Different inputs (x) yieldthe same output (y).

The slope is not equal to zero The slope is equal to zero

Linear relationship.Linear relationship.Linear relationship.Linear relationship.

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• We can draw inference about b1 from b1 by testingH0: b1 = 0H1: b1 = 0 (or < 0,or > 0)– The test statistic is

– If the error variable is normally distributed, the statistic is Student t distribution with d.f. = n-2.

1b

11

sb

tb

The standard error of b1.

1 1bx

ss

n se

where

Testing the Slope

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• Example– Test to determine whether there is enough evidence

to infer that there is a linear relationship between the car auction price and the odometer reading for all three-year-old Tauruses in the previous example . Use a = 5%.

Testing the Slope,Example

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• Solving by hand– To compute “t” we need the values of b1 and sb1.

– The rejection region is t > t.025 or t < -t.025 with n = n-2 = 98, t.025

= 1.9845

1

1

1

1 1

.06686326.49 .004975

1 99 6596.125

06686 0 13.44004975

bx

b

bssn s

b .t .s

e

b

Testing the Slope,Example

Odometer Price37400 1460044800 14100 Regression Statistics45800 14000 Multiple R 0.80516797930900 15600 R Square 0.648295475

31700 15600Adjusted R Square 0.644706653

34000 14700Standard Error 326.4886258

45900 14500Observations 100

19100 1570040100 15100 ANOVA40200 14800 df SS MS F Significance F

32400 15200 Regression 1 19255607.37 19255607.4 180.643 5.75078E-2443500 14700 Residual 98 10446292.63 106594.82332700 15600 Total 99 29701900 34500 15600

37700 14600 Coefficients Standard Error t Stat P-value Lower 95% Upper 95%41400 14600 Intercept 17248.72734 182.0925742 94.7250453 3.57E-98 16887.37056 17610.0824500 15700 Odometer -0.066860885 0.004974639 -13.4403493 5.75E-24 -0.076732895 -0.0569935800 1500048600 1470024200 15400

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• Using the computer

There is overwhelming evidence to inferthat the odometer reading affects the auction selling price.

Testing the Slope (Example)

Coefficient of determination

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Reduction in prediction error when use x:TSS-SSE = SSR

10

2 2

1 1

2

1

Case I: Case II:

ˆignore : use to predict use : use

errors: (obs. pred.) errors: (obs. pred.)

ˆ ( ) = (

n n

i i

n

i ii

x y y x y b b x

y y y

+

2

1

)

n

ii

y

TSS SSE


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Reduction in prediction error when use x:TSS-SSE = SSR or TSS = SSR + SSE

2

212 2

Proportional reduction in prediction error when use :

( )( )1 algebra =

n

i ii

x y

x

x x y yTSS SSE SSE rTSS TSS s s

Overall variability in yTSS

The regression modelSSR

Remains, in part, unexplainedThe error

SSE

Explained in part by

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Coefficient of determination: graphically

x1 x2

y1

y2

y

Two data points (x1,y1) and (x2,y2) of a certain sample are shown.

+ 22

21 )yy()yy( 2

22

1 )yy()yy( + 222

211 )yy()yy( ++

Total variation in y = Variation explained by the regression line

+ Unexplained variation (error)

Variation in y = SSR + SSE (TSS)

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Coefficient of determination• R2 (=r2 ) measures the proportion of the variation

in y that is explained by the variation in x.

2 1 SSE TSS SSE SSRRTSS TSS TSS

• r2 takes on any value between zero and one (-1r 1).r2 = 1: Perfect match between the line and the data points.r2 = 0: There are no linear relationship between x and y.

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• Example– Find the coefficient of determination for the used car

price –odometer example.what does this statistic tell you about the model?

• Solution– Solving by hand;

2 2( .80517) .6483r

Coefficient of determination,Example

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– Using the computer From the regression output we have


Regression StatisticsMultiple R 0.805167979R Square 0.648295475Adjusted R Square 0.644706653Standard Error 326.4886258Observations 100

ANOVA df SS MS F Significance F

Regression 1 19255607.37 19255607.37 180.643 5.75078E-24Residual 98 10446292.63 106594.8228Total 99 29701900

Coefficients Standard Error t Stat P-valueIntercept 17248.72734 182.0925742 94.72504534 3.57E-98Odometer -0.06686089 0.004974639 -13.44034928 5.75E-24

64.8% of the variation in the auctionselling price is explained by the variation in odometer reading. Therest (35.2%) remains unexplained bythis model.

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• If we are satisfied with how well the model fits the data, we can use it to predict the values of y.

• To make a prediction we use– Point prediction, and– Interval prediction

Using the Regression Equation

• Before using the regression model, we need to assess how well it fits the data.

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Point Prediction• Example

– Predict the selling price of a three-year-old Taurus with 40,000 miles on the odometer.

– It is predicted that a 40,000 miles car would sell for $14,574.

– How close is this prediction to the real price?

ˆ 17248.73 .06686 17248.73 .066686(40,000) 14,574y x A point prediction

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Interval Estimates• Two intervals can be used to discover how closely the

predicted value will match the true value of y.– Prediction interval – predicts y for a given value of x,– Confidence interval – estimates the average y for a given x.

– The confidence interval

– The prediction interval2

2 2 22 1ˆ ( )( ) sy t SE b x x s

ne

a n e + +

22 2

2 1ˆ ( )( ) sy t SE b x xne

a n +

35

Interval Estimates,Example

• Example - continued – Provide an interval estimate for the bidding price on

a Ford Taurus with 40,000 miles on the odometer.– Two types of predictions are required:

• A prediction for a specific car• An estimate for the average price per car

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• Solution– A prediction interval provides the price estimate for a

single car:

22 2 2326.4914,574 1.9845 (.004975) (40,000 36,011) 326.49 14,574 652

100 + +

t.025,98

22 2 2

2 1ˆ ( )( ) sy t SE b x x sne

a n e + +

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• Solution – continued– A confidence interval provides the estimate of the

mean price per car for a Ford Taurus with 40,000 miles reading on the odometer.

• The confidence interval (95%) =


22 2

2 1ˆ ( )( ) sy t SE b x x

ne

a n +

22 2 326.4914,574 1.9845 (.004975) (40,000 36,011) 14,574 76

100 +

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– As x moves away from x the interval becomes longer. That is, the shortest interval is found at x.

22 2

2 1ˆ ( ) ( ) sy t SE b x xne

a n +

x

0 1y b b xn +

The effect of the given x on the length of the interval

22 2

2 1ˆ ( )(1) sy t SE bne

a +

39

x1x)1x( 1x)1x( +

0 1y b b xn +

ˆ( 1)y x xn

ˆ( 1)y x xn +

1x +1x



22 2

2 1ˆ ( ) ( ) sy t SE b x xne

a n +

40

x


0 1y b b xn +

2x)2x( 2x)2x( +

2x 2x +


22 2

2 1ˆ ( ) ( ) sy t SE b x xne

a n +

22 2

2 1ˆ ( )(1) sy t SE bne

a +

22 2

2 1ˆ ( )(2) sy t SE bne

a +

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Regression Diagnostics - I

• The three conditions required for the validity of the regression analysis are:– the error variable is normally distributed.– the error variance is constant for all values of x.– The errors are independent of each other.

• How can we diagnose violations of these conditions?

42

Residual Analysis• Examining the residuals (or standardized

residuals), help detect violations of the required conditions.

• Example – continued:– Nonnormality.

• Use Excel to obtain the standardized residual histogram.• Examine the histogram and look for a bell shaped.

diagram with a mean close to zero.

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For each residual we calculate the standard deviation as follows:

2x

2i

i

ir

s)1n()xx(

n1

h

whereh1ssi

+

e

A Partial list ofStandard residuals

ObservationPredicted Price Residuals Standard Residuals1 14736.91 -100.91 -0.332 14277.65 -155.65 -0.523 14210.66 -194.66 -0.654 15143.59 446.41 1.485 15091.05 476.95 1.58

Standardized residual ‘i’ =Residual ‘i’

Standard deviation

Residual Analysis

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Standardized residuals

0

10

20

30

40

-2 -1 0 1 2 More

It seems the residual are normally distributed with mean zero

Residual Analysis

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Heteroscedasticity• When the requirement of a constant variance is violated we have

a condition of heteroscedasticity.• Diagnose heteroscedasticity by plotting the residual against the

predicted y.

+ + ++

+ ++

++

+

+

+

+

+

+

+

+

+

+

++

+

+

+

The spread increases with y

y

Residualy

+

+++

+

++

+

++

+

+++

+

+

+

+

+

++

+

+

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Homoscedasticity• When the requirement of a constant variance is not

violated we have a condition of homoscedasticity.• Example - continued

-1000

-500

0

500

1000

13500 14000 14500 15000 15500 16000

Predicted Price

Resi

dual

s

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Non Independence of Error Variables– A time series is constituted if data were collected

over time.– Examining the residuals over time, no pattern should

be observed if the errors are independent.– When a pattern is detected, the errors are said to be

autocorrelated.– Autocorrelation can be detected by graphing the

residuals against time.

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Patterns in the appearance of the residuals over time indicates that autocorrelation exists.

+

+++ +

++

++ + +

++ + +

++ + +

+

+

+

+

+

+Time

Residual Residual

Time+

+

+

Note the runs of positive residuals,replaced by runs of negative residuals

Note the oscillating behavior of the residuals around zero.

0 0

Non Independence of Error Variables

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Outliers• An outlier is an observation that is unusually small or large.• Several possibilities need to be investigated when an outlier

is observed:– There was an error in recording the value.– The point does not belong in the sample.– The observation is valid.

• Identify outliers from the scatter diagram.• It is customary to suspect an observation is an outlier if its |

standard residual| > 2

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+

+

+

++ +

+ + ++

++

+

+

+

+

+

The outlier causes a shift in the regression line

… but, some outliers may be very influential

++++++++++

An outlier An influential observation

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Procedure for Regression Diagnostics• Develop a model that has a theoretical basis.• Gather data for the two variables in the model.• Draw the scatter diagram to determine whether a linear model

appears to be appropriate.• Determine the regression equation.• Check the required conditions for the errors.• Check the existence of outliers and influential observations• Assess the model fit.• If the model fits the data, use the regression equation.

Simple Linear Regression 1. the least squares procedure 2. inference for least squares lines

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