04/22/23 http://numericalmethods.eng.usf.edu 1

Adequacy of Linear Regression Models

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

Data

-350 -300 -250 -200 -150 -100 -50 0 50 1002

2.5

3

3.5

4

4.5

5

5.5

6

6.5

x

yy vs x

Is this adequate?

-350 -300 -250 -200 -150 -100 -50 0 50 1002

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

x

yy vs x

Straight Line Model

Quality of Fitted Data• Does the model describe the data

adequately?

• How well does the model predict the response variable predictably?

Linear Regression Models• Limit our discussion to adequacy of

straight-line regression models

Four checks

1. Plot the data and the model.2. Find standard error of estimate.3. Calculate the coefficient of

determination.4. Check if the model meets the

assumption of random errors.

Example: Check the adequacy of the straight line model for given data

T(F)

α (μin/in/F)

-340 2.45

-260 3.58

-180 4.52

-100 5.28

-20 5.86

60 6.36

Taa 10

END

1. Plot the data and the model

Data and model

T(F)

α (μin/in/F)

-340 2.45-260 3.58-180 4.52-100 5.28-20 5.8660 6.36

-350 -300 -250 -200 -150 -100 -50 0 50 1002

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

T

TT 0096964.00325.6)(

END

2. Find the standard error of estimate

Standard error of estimate

2/

nSs r

T

n

iiir TaaS

1

210 )(

Standard Error of Estimate

-340-260-180-100-2060

2.453.584.525.285.866.36

2.73573.51144.28715.06295.83866.6143

-0.285710.0685710.232860.21714

0.021429-0.25429

iT i iTaa 10 ii Taa 10

TT 0096964.00325.6)(

Standard Error of Estimate

25283.0rS

2/

nS

s rT

2625283.0

25141.0

Standard Error of Estimate

T

ii

sTaa

/

10 ResidualScaled

-350 -300 -250 -200 -150 -100 -50 0 50 1002

3

4

5

6

7

8

T

Scaled Residuals

T

ii

sTaa

/

10 ResidualScaled

Estimateof Error StandardResidual ResidualScaled

95% of the scaled residuals need to be in [-2,2]

Scaled Residuals

Ti αi Residual Scaled Residual

-340-260-180-100-2060

2.453.584.525.285.866.36

-0.285710.0685710.232860.217140.021429-0.25429

-1.13640.272750.926220.86369

0.085235-1.0115

25141.0/ Ts

END

3. Find the coefficient of determination

Coefficient of determination

n

iiir TaaS

1

210

n

iitS

1

2

t

rt

SSS

r

2

Sum of square of residuals between data and mean

n

iit yyS

1

2

11, yx 33 , yx

22 , yx

),( nn yx

ii yx ,y

x

_

yy _

yy

Sum of square of residuals between observed and predicted

n

iiir xaayS

1

210

11, yx

33 , yx

22 , yx

),( nn yx ii yx ,

iii xaayE 10

y

x

Limits of Coefficient of Determination

t

rt

SSS

r

2

10 2 r

Calculation of St

-340-260-180-100-2060

2.453.584.525.285.866.36

-2.2250-1.09500.155000.605001.18501.6850

iT i i

783.10tS6750.4

Calculation of Sr

-340-260-180-100-2060

2.453.584.525.285.866.36

2.73573.51144.28715.06295.83866.6143

-0.285710.0685710.232860.21714

0.021429-0.25429

iT i iTaa 10 ii Taa 10

25283.0rS

Coefficient of determination

t

rt

SSS

r

2

783.1025283.0783.10

97655.0

Correlation coefficient

t

rt

SSS

r

98820.0

How do you know if r is positive or negative ?

What does a particular value of |r| mean?

0.8 to 1.0 - Very strong relationship 0.6 to 0.8 - Strong relationship 0.4 to 0.6 - Moderate relationship 0.2 to 0.4 - Weak relationship 0.0 to 0.2 - Weak or no relationship 

Caution in use of r2

• Increase in spread of regressor variable (x) in y vs. x increases r2

• Large regression slope artificially yields high r2

• Large r2 does not measure appropriateness of the linear model

• Large r2 does not imply regression model will predict accurately

Final Exam Grade

Final Exam Grade vs Pre-Req GPA

END

4. Model meets assumption of random errors

Model meets assumption of random errors

• Residuals are negative as well as positive

• Variation of residuals as a function of the independent variable is random

• Residuals follow a normal distribution• There is no autocorrelation between the

data points.

Therm exp coeff vs temperatureT α60 6.3640 6.2420 6.120 6.00-20 5.86-40 5.72-60 5.58-80 5.43

T α-100 5.28-120 5.09-140 4.91-160 4.72-180 4.52-200 4.30-220 4.08-240 3.83

T α-280 3.33-300 3.07-320 2.76-340 2.45

Data and model

-350 -300 -250 -200 -150 -100 -50 0 50 1002

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

T

T0093868.00248.6

Plot of Residuals

-350 -300 -250 -200 -150 -100 -50 0 50 100-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

T

Resid

ual

Histograms of Residuals

Check for Autocorrelation• Find the number of times, q the sign of the

residual changes for the n data points.• If (n-1)/2-√(n-1) ≤q≤ (n-1)/2+√(n-1), you

most likely do not have an autocorrelation.

1222

1221222

)122(

q

083.159174.5 q

Is there autocorrelation?

083.159174.5 q

-350 -300 -250 -200 -150 -100 -50 0 50 100-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

T

Resid

ual

y vs x fit and residuals

n=40

Is 13.3≤21≤ 25.7? Yes!(n-1)/2-√(n-1) ≤p≤ (n-1)/2+√(n-1)

y vs x fit and residuals

(n-1)/2-√(n-1) ≤p≤ (n-1)/2+√(n-1)Is 13.3≤2≤ 25.7? No!

n=40

END

What polynomial model to choose if one needs to be chosen?

First Order of Polynomial

-350 -300 -250 -200 -150 -100 -50 0 50 1002

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7x 10

-6 Polynomial Regression of order 1

x

y =

a 0+a1*x

+a2*x

2 +....

.+a m

*xm

Second Order Polynomial

-350 -300 -250 -200 -150 -100 -50 0 50 1002

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7x 10

-6 Polynomial Regression of order 2

x

y =

a 0+a1*x

+a2*x

2 +....

.+a m

*xm

Which model to choose?

-350 -300 -250 -200 -150 -100 -50 0 50 1002

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

x

yy vs x

Optimum Polynomial

0 1 2 3 4 5 60

1

2

3

4

5

x 10-14 Optimum Order of Polynomial

Order of Polynomial, m

Sr

[n-(m

+1)]

THE END

Effect of an Outlier

Effect of Outlier

y = 2xR2 = 1

0

5

10

15

20

25

0 2 4 6 8 10 12

Effect of Outlier

y = 3.2727x - 5.0909R2 = 0.6879

-10

0

10

20

30

40

50

60

0 2 4 6 8 10 12