Simple lin regress_inference

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Simple Linear RegressionSimple Linear Regression1. review of least squares 1. review of least squares

procedureprocedure2. inference for least squares lines 2. inference for least squares lines

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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 dependent variable (y) from

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

independent variables and the dependent variable.

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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 components

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House cost = 25000 + 75(Size)

House size

HouseCost

Most lots sell for $25,000

However, house cost vary even among same size houses!

The Model

Since cost behave unpredictably, we add a random component.

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The Model

• The first order linear model

y = dependent variablex = independent variable0 = y-intercept1 = slope of the line = error variable

xy 10 xy 10

x

y

0 Run

Rise = Rise/Run

0 and 1 are unknown populationparameters, therefore are estimated from the data.

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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.

Question: What should be considered a good line?

x

y

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The Least Squares (Regression) Line

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

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The Least Squares (Regression) Line

3

3

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.

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The Estimated Coefficients

To calculate the estimates of the slope and intercept of the least squares line , use the formulas:

1

0 1

2

2 2( 1)

xy

xx

i i

xy i i

i

xx i x

SSb

SS

b y b x

x ySS x y

n

xSS x n s

n

1

0 1

2

2 2( 1)

xy

xx

i i

xy i i

i

xx i x

SSb

SS

b y b x

x ySS x y

n

xSS x n s

n

The regression equation that estimatesthe equation of the first order linear modelis:

0 1y b b x 0 1y b b x

1y

x

sb r

s

1y

x

sb r

s

Alternate formula for the slope b1

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• 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 37388 146362 44758 141223 45833 140164 30862 155905 31705 155686 34010 14718

. . .

. . .

. . .

Independent variable x

Dependent variable y

The Simple Linear Regression Line

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• Solution– Solving by hand: Calculate a number of statistics

;823.822,14y

;45.009,36x

22 43,528,690

( ) 2,712,511

i

xx i

i ixy i i

xSS x

n

x ySS x y

n

where n = 100.

1 2

0 1

2,712,511.06232

( 1) 43,528,690

14,822.82 ( .06232)(36,009.45) 17,067

xy

x

SSb

n s

b y b x

x0623.067,17xbby 10

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• Solution – continued– Using the computer

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


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SUMMARY OUTPUT

Regression StatisticsMultiple R 0.8063R Square 0.6501Adjusted R Square0.6466Standard Error 303.1Observations 100

ANOVAdf SS MS F Significance F

Regression 1 16734111 16734111 182.11 0.0000Residual 98 9005450 91892Total 99 25739561

CoefficientsStandard Error t Stat P-valueIntercept 17067 169 100.97 0.0000Odometer -0.0623 0.0046 -13.49 0.0000

xy 0623.067,17ˆ


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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 $0.0623

Odometer Line Fit Plot

13000

14000

15000

16000

Odometer

Pri

ce

xy 0623.067,17ˆ

Interpreting the Linear Regression -Equation

The intercept is b0 = $17067.

0 No data

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

17067

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Error Variable: Required Conditions

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

be satisfied.– The probability distribution of is normal.– The mean of is zero: E() = 0.– The standard deviation of is 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

From the first three assumptions we have:y is normally distributed with meanE(y) = 0 + 1x, and a constant standard deviation

From the first three assumptions we have:y is normally distributed with meanE(y) = 0 + 1x, and a constant standard deviation

0 + 1x1

0 + 1x2

0 + 1x3

E(y|x2)

E(y|x3)

x1 x2 x3

E(y|x1)

The standard deviation remains constant,

but the mean value changes with x

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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

.)yy(SSEn

1i

2ii

Sum of Squares for Errors

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

– A shortcut formula

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– The mean error is equal to zero.– If 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 as a measure of the suitability of using a linear model.

– An estimator of is given by s

2

tan

n

SSEs

EstimateofErrordardS

2

tan

n

SSEs

EstimateofErrordardS

Standard Error of Estimate

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

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

9,005,450

9,005,450303.13

2 98

SSE

SSEs

n

It is hard to assess the model based

on s even when compared with the mean value of y.

823,14y1.303s

Standard Error of Estimate,Example

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Testing the slope– When no linear relationship exists between two

variables, the regression line should be horizontal.

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 1 from b1 by testingH0: 1 = 0H1: 1 = 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

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sb

t

1b

11

sb

t

The standard error of b1.

1b

xx

ss

SS

1b

xx

ss

SSwhere

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 = 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-2 = 98.Approximately, t.025 = 1.984

49.1300462

00623

00462.)690,528,43)(99(

1.303

)1(

0623.

1

1

11

2

1

.

.s

bt

sn

ss

b

b

x

b


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Price Odometer SUMMARY OUTPUT14636 3738814122 44758 Regression Statistics14016 45833 Multiple R 0.806315590 30862 R Square 0.650115568 31705 Adjusted R Square0.646614718 34010 Standard Error 303.114470 45854 Observations 10015690 1905715072 40149 ANOVA14802 40237 df SS MS F Significance F15190 32359 Regression 1 16734111 16734111 182.11 0.000014660 43533 Residual 98 9005450 9189215612 32744 Total 99 2573956115610 3447014634 37720 CoefficientsStandard Error t Stat P-value14632 41350 Intercept 17067 169 100.97 0.000015740 24469 Odometer -0.0623 0.0046 -13.49 0.0000

• Using the computer

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


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– To measure the strength of the linear relationship we use the coefficient of determination.

2

22 2

22

( )(

1( )

i i

x y

i

x x y yR

s s

SSEor R

y y

2

22 2

22

( )(

1( )

i i

x y

i

x x y yR

s s

SSEor R

y y

Coefficient of determination

Note that the coefficient of determination is r2

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• To understand the significance of this coefficient note:

Overall variability in yThe regression model

Remains, in part, unexplained The error

Explained in part by

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x1 x2

y1

y2

y

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

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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

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• R2 measures the proportion of the variation in y that is explained by the variation in x.

2i

2i

2i

2i

2

)yy(

SSR

)yy(

SSE)yy(

)yy(

SSE1R

• R2 takes on any value between zero and one.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

[ 2,712,511]2(43,528,688)(259,996)2 2

( )(.6501i i

x y

x x y yR

s s

Coefficient of determination,Example

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SUMMARY OUTPUT

Regression StatisticsMultiple R 0.8063R Square 0.6501Adjusted R Square0.6466Standard Error 303.1Observations 100

ANOVAdf SS MS F Significance F

Regression 1 16734111 16734111 182.11 0.0000Residual 98 9005450 91892Total 99 25739561

CoefficientsStandard Error t Stat P-valueIntercept 17067 169 100.97 0.0000Odometer -0.0623 0.0046 -13.49 0.0000

– Using the computer From the regression output we have

65% of the variation in the auctionselling price is explained by the variation in odometer reading. Therest (35%) 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 $14,575.

– How close is this prediction to the real price?

575,14)000,40(0623.17067x0623.17067y 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 confidence interval2

2 2

( )1ˆ

( )

g

i

x xy t s

n x x

2

2 2

( )1ˆ

( )

g

i

x xy t s

n x x

– The prediction interval– The prediction interval2

2 2

( )1ˆ 1

( )

g

i

x xy t s

n x x

2

2 2

( )1ˆ 1

( )

g

i

x xy t s

n x x

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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: 2

2 2

( )1ˆ 1

( )

g

i

x xy t s

n x x

21 (40,000 36,009)[17,067 .0623(40000)] 1.984(303.1) 1 14,575 605

100 4,309,340,310

t.025,98

Approximately

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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%) =

2

i

2g

2)xx(

)xx(

n1

sty

21 (40,000 36,009)[17,067 .0623(40000)] 1.984(303.1) 14,575 70

100 4,309,340,310


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

2x

2g

2 s)1n(

)xx(

n1

sty

x

g10 xbby

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

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x1x)1x( 1x)1x(

g10 xbby

)1xx(y g )1xx(y g

1x 1x



2x

2g

2 s)1n(

)xx(

n1

sty

2x

2

2 s)1n(1

n1

sty

40

x


g10 xbby

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

2x 2x

2x

2g

2 s)1n()xx(

n1

sty

2x

2

2 s)1n(1

n1

sty

2x

2

2 s)1n(2

n1

sty


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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?

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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

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

Re

sid

ua

ls

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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 lin regress_inference

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