regresi 1. 2 regresi adalah diberikan sejumlah n buah data yg dimodelkan oleh persamaan model yg...

REGRESI

1

2

Regresi adalah

Diberikan sejumlah n buah data ),( , ... ),,(),,( 2211 nn yxyx yxYg dimodelkan oleh persamaan

)(xfy Model yg paling baik (best fit) secara umum adalah model yg meminimalkan jumlah kuadrat residual

rS

)( iii xfy

n

i

iir xfyS1

2))((),( 11 yx

),( nn yx

)(xfy

Figure. Basic model for regression

Jumlah kuadrat residual

3

REGRESI LINIER

4

Regresi Linier (Kriteria 1)

),( , ... ),,(),,( 2211 nn yxyx yx

Diberikan sejumlah n data xaay 10 Best fit dimodelkan dalam bentuk persamaan

x

iiixaay

10

11, yx

22, yx

33, yx

nnyx ,

iiyx ,

iiixaay

10

y

Figure. Linear regression of y vs. x data showing residuals at a typical

point, xi .

5

Contoh Kriteria 1

x y

2.0 4.0

3.0 6.0

2.0 6.0

3.0 8.0

Diberikan sejumlah titik (2,4), (3,6), (2,6) and (3,8), best fit dimodelkan dalam bentuk persamaan garis lurus

Figure. Data points for y vs. x data.

Table. Data Points

0

2

4

6

8

10

0 1 2 3 4

x

y

6

04

1

i

i

x y ypredicted ε = y - ypredicted

2.0 4.0 4.0 0.0

3.0 6.0 8.0 -2.0

2.0 6.0 4.0 2.0

3.0 8.0 8.0 0.0

Table. Residuals at each point for regression model y = 4x – 4.

Figure. Regression curve for y=4x-4, y vs. x data

0

2

4

6

8

10

0 1 2 3 4

x

y

Dengan menggunakan persamaan y=4x-4 maka diperoleh kurva regresi

7

x y ypredicted ε = y - ypredicted

2.0 4.0 6.0 -2.0

3.0 6.0 6.0 0.0

2.0 6.0 6.0 0.0

3.0 8.0 6.0 2.0

04

1

i

i

0

2

4

6

8

10

0 1 2 3 4

x

y

Table. Residuals at each point for y=6

Figure. Regression curve for y=6, y vs. x data

Persamaan y=6

8

04

1

i

i Kedua persamaan y=4x-4 and y=6 memiliki residual minimum tetapi memiliki model regresi yang tidak unik. Oleh karena itu kriteria 1 merupakan kriteria yang buruk

9

Regresi linier (Kriteria 2)

x

iiixaay

10

11, yx

22, yx

33, yx

nnyx ,

iiyx ,

iiixaay

10

y


point, xi .

Meminimalkan dengan memberikan harga mutlak

n

ii

1

10

x y ypredicted |ε| = |y - ypredicted|

2.0 4.0 4.0 0.0

3.0 6.0 8.0 2.0

2.0 6.0 4.0 2.0

3.0 8.0 8.0 0.0

0

2

4

6

8

10

0 1 2 3 4

x

y

Table. The absolute residuals employing the y=4x-4 regression model

Figure. Regression curve for y=4x-4, y vs. x data

44

1

i

i

Dengan menggunakan persamaan y=4x-4

11

x y ypredicted |ε| = |y – ypredicted|

2.0 4.0 6.0 2.0

3.0 6.0 6.0 0.0

2.0 6.0 6.0 0.0

3.0 8.0 6.0 2.0

44

1

i

i

Table. Absolute residuals employing the y=6 model

0

2

4

6

8

10

0 1 2 3 4

xy

Figure. Regression curve for y=6, y vs. x data

Dengan persamaan y=6

12

Can you find a regression line for which

44

1

i

i and has unique

regression coefficients?

for both regression models of y=4x-4 and y=6.

The sum of the errors has been made as small as possible, that is 4, but the regression model is not unique.

Hence the above criterion of minimizing the sum of the absolute value of the residuals is also a bad criterion.

44

1

i

i

13

Least Squares Criterion

Kriteria Least Squares meminimalkan jumlah kuadrat residual dari model Persamaan.

2

110

1

2

n

iii

n

iir xaayS

x

iiixaay

10

11, yx

22, yx

33, yx

nnyx ,

iiyx ,

iiixaay

10

y


point, xi .

14

Finding Constants of Linear Model

2

110

1

2

n

iii

n

iir xaayS Minimize the sum of the square of the

residuals:To find

0121

100

n

iii

r xaaya

S

021

101

n

iiii

r xxaaya

S

giving

i

n

iii

n

ii

n

i

xyxaxa

1

2

11

10

0a and

1a we minimize

with respect to

1a 0aand

rS .

n

iii

n

i

n

i

yxaa11

11

0

)( 10 xaya

15

Finding Constants of Linear Model

0a

Solving for

2

11

2

1111

n

ii

n

ii

n

ii

n

ii

n

iii

xxn

yxyxn

a

and

2

11

2

1111

2

0

n

ii

n

ii

n

iii

n

ii

n

ii

n

ii

xxn

yxxyx

a

1aand directly yields,

)( 10 xaya

16

Example 1

The torque, T needed to turn the torsion spring of a mousetrap through an angle, is given below.

Angle, θ Torque, T

Radians N-m

0.698132 0.188224

0.959931 0.209138

1.134464 0.230052

1.570796 0.250965

1.919862 0.313707

Table: Torque vs Angle for a torsional spring

Find the constants for the model given by21 kkT

Figure. Data points for Angle vs. Torque data

0.1

0.2

0.3

0.4

0.5 1 1.5 2

θ (radians)

To

rqu

e (

N-m

)

17

Example 1 cont.

1a

The following table shows the summations needed for the calculations of the constants in the regression model.

2 T

Radians N-m Radians2 N-m-Radians

0.698132 0.188224 0.487388 0.131405

0.959931 0.209138 0.921468 0.200758

1.134464 0.230052 1.2870 0.260986

1.570796 0.250965 2.4674 0.394215

1.919862 0.313707 3.6859 0.602274

6.2831 1.1921 8.8491 1.5896

Table. Tabulation of data for calculation of important

5

1i

5nUsing equations described for

25

1

5

1

2

5

1

5

1

5

12

ii

ii

ii

ii

iii

n

TTnk

228316849185

1921128316589615

..

...

21060919 . N-m/rad

summations

0aTand

with

18

Example 1 cont.

n

TT i

i

5

1_

Use the average torque and average angle to calculate

1k

_

2

_

1 kTk

ni

i

5

1_

5

1921.1

1103842.2

5

2831.6

2566.1

Using,

)2566.1)(106091.9(103842.2 21 1101767.1 N-m

19

Example 1 Results

Figure. Linear regression of Torque versus Angle data

Using linear regression, a trend line is found from the data

Can you find the energy in the spring if it is twisted from 0 to 180 degrees?

20

Example 2

Strain Stress

(%) (MPa)

0 0

0.183 306

0.36 612

0.5324 917

0.702 1223

0.867 1529

1.0244 1835

1.1774 2140

1.329 2446

1.479 2752

1.5 2767

1.56 2896

To find the longitudinal modulus of composite, the following data is collected. Find the longitudinal modulus, Table. Stress vs. Strain data

E using the regression model E and the sum of the square of

the

0.0E+00

1.0E+09

2.0E+09

3.0E+09

0 0.005 0.01 0.015 0.02

Strain, ε (m/m)

Str

ess,

σ (

Pa)

residuals.

Figure. Data points for Stress vs. Strain data

21

Example 2 cont.

iii E Residual at each point is given by

The sum of the square of the residuals then is

n

iirS

1

2

n

iii E

1

2

0)(21

i

n

iii

r EE

S

Differentiate with respect to

E

n

ii

n

iii

E

1

2

1

Therefore

22

Example 2 cont.

i ε σ ε 2 εσ

1 0.0000 0.0000 0.0000 0.0000

2 1.8300×10−3 3.0600×108 3.3489×10−6 5.5998×105

3 3.6000×10−3 6.1200×108 1.2960×10−5 2.2032×106

4 5.3240×10−3 9.1700×108 2.8345×10−5 4.8821×106

5 7.0200×10−3 1.2230×109 4.9280×10−5 8.5855×106

6 8.6700×10−3 1.5290×109 7.5169×10−5 1.3256×107

7 1.0244×10−2 1.8350×109 1.0494×10−4 1.8798×107

8 1.1774×10−2 2.1400×109 1.3863×10−4 2.5196×107

9 1.3290×10−2 2.4460×109 1.7662×10−4 3.2507×107

10 1.4790×10−2 2.7520×109 2.1874×10−4 4.0702×107

11 1.5000×10−2 2.7670×109 2.2500×10−4 4.1505×107

12 1.5600×10−2 2.8960×109 2.4336×10−4 4.5178×107

1.2764×10−3 2.3337×108

Table. Summation data for regression model

12

1i

12

1

32 102764.1i

i

With

and

12

1

8103337.2i

ii

Using

12

1

2

12

1

ii

iii

E

3

8

102764.1

103337.2

GPa84.182

23

Example 2 Results 84.182The

equation

Figure. Linear regression for Stress vs. Strain data

describes the data.

REGRESI NON LINIER

Nonlinear Regression

)( bxaey

)( baxy

xb

axy

Some popular nonlinear regression models:

1. Exponential model:2. Power model:

3. Saturation growth model:4. Polynomial model: )( 10

mmxa...xaay

25

Nonlinear Regression

Given n data points

),( , ... ),,(),,( 2211 nn yxyx yx best fit )(xfy

to the data, where

)(xf is a nonlinear function of

x .

Figure. Nonlinear regression model for discrete y vs. x data

)(xfy

),(nn

yx

),(11

yx

),(22

yx

),(ii

yx

)(ii

xfy

26

RegressionExponential Model

27

Exponential Model),( , ... ),,(),,( 2211 nn yxyx yxGive

nbest fit

bxaey to the data.

Figure. Exponential model of nonlinear regression for y vs. x data

bxaey

),(nn

yx

),(11

yx

),(22

yx

),(ii

yx

)(ii

xfy

28

Finding Constants of Exponential Model

n

i

bx

ir iaeyS

1

2

The sum of the square of the residuals is defined as

Differentiate with respect to a and b

021

ii bxn

i

bxi

r eaeya

S

021

ii bxi

n

i

bxi

r eaxaeyb

S

29

Finding Constants of Exponential Model

Rewriting the equations, we obtain

01

2

1

n

i

bxn

i

bxi

ii eaey

01

2

1

n

i

bxi

n

i

bxii

ii exaexy

30

Finding constants of Exponential Model

Substituting a back into the previous equation

01

2

1

2

1

1

n

i

bxin

i

bx

bxn

ii

bxi

n

ii

i

i

i

i exe

eyexy

The constant b can be found through numerical methods such as bisection method.

n

i

bx

n

i

bxi

i

i

e

eya

1

2

1

Solving the first equation for a yields

31

Example 1-Exponential Model

t(hrs) 0 1 3 5 7 9

1.000 0.891 0.708 0.562 0.447 0.355

Many patients get concerned when a test involves injection of a radioactive material. For example for scanning a gallbladder, a few drops of Technetium-99m isotope is used. Half of the techritium-99m would be gone in about 6 hours. It, however, takes about 24 hours for the radiation levels to reach what we are exposed to in day-to-day activities. Below is given the relative intensity of radiation as a function of time.

Table. Relative intensity of radiation as a function of time.

32

Example 1-Exponential Model cont.

Find: a) The value of the regression

constants A an

db) The half-life of Technium-99m

c) Radiation intensity after 24 hours

The relative intensity is related to time by the equation tAe

33

Plot of data

34

Constants of the Model

The value of λ is found by solving the nonlinear equation

01

2

1

2

1

1

n

i

tin

i

t

n

i

ti

ti

n

ii

i

i

i

i ete

eetf

n

i

t

n

i

ti

i

i

e

eA

1

2

1

tAe

35

Setting up the Equation in MATLAB

01

2

1

2

1

1

n

i

tin

i

t

n

i

ti

ti

n

ii

i

i

i

i ete

eetf

t (hrs) 0 1 3 5 7 9

γ 1.000

0.891

0.708

0.562

0.447

0.35536

Setting up the Equation in MATLAB

01

2

1

2

1

1

n

i

tin

i

t

n

i

ti

ti

n

ii

i

i

i

i ete

eetf

t=[0 1 3 5 7 9]gamma=[1 0.891 0.708 0.562 0.447 0.355]syms lamdasum1=sum(gamma.*t.*exp(lamda*t));sum2=sum(gamma.*exp(lamda*t));sum3=sum(exp(2*lamda*t));sum4=sum(t.*exp(2*lamda*t));f=sum1-sum2/sum3*sum4;

1151.0

37

Calculating the Other Constant

The value of A can now be calculated

6

1

2

6

1

i

t

i

ti

i

i

e

eA

9998.0

The exponential regression model then is te 1151.0 9998.0

38

Plot of data and regression curve

te 1151.0 9998.0

39

Relative Intensity After 24 hrs

The relative intensity of radiation after 24 hours 241151.09998.0 e

2103160.6 This result implies that only

%317.61009998.0

10316.6 2

radioactive intensity is left after 24 hours.

40

Homework• What is the half-life of technetium

99m isotope?• Compare the constants of this

regression model with the one where the data is transformed.

• Write a program in the language of your choice to find the constants of the model.

41

Polynomial Model),( , ... ),,(),,( 2211 nn yxyx yxGiven best

fit

m

mxa...xaay

10

)2( nm to a given data set.

Figure. Polynomial model for nonlinear regression of y vs. x data

m

mxaxaay

10

),(nn

yx

),(11

yx

),(22

yx

),(ii

yx

)(ii

xfy

42

Polynomial Model cont.The residual at each data point is given by

mimiii xaxaayE ...10

The sum of the square of the residuals then is

n

i

mimii

n

iir

xaxaay

ES

1

2

10

1

2

...

43

Polynomial Model cont.To find the constants of the polynomial model, we set the derivatives with respect to ia wher

e

0)(....2

0)(....2

0)1(....2

110

110

1

110

0

mi

n

i

mimii

m

r

i

n

i

mimii

r

n

i

mimii

r

xxaxaaya

S

xxaxaaya

S

xaxaaya

S

,,1 mi equal to zero.

44

Polynomial Model cont.These equations in matrix form are

given by

n

ii

mi

n

iii

n

ii

mn

i

mi

n

i

mi

n

i

mi

n

i

mi

n

ii

n

ii

n

i

mi

n

ii

yx

yx

y

a

a

a

xxx

xxx

xxn

1

1

1

1

0

1

2

1

1

1

1

1

1

2

1

11

......

...

...........

...

...

The above equations are then solved for

maaa ,,, 10

45

Example 2-Polynomial Model

Temperature, T(oF)

Coefficient of thermal

expansion, α (in/in/oF)

80 6.47×10−6

40 6.24×10−6

−40 5.72×10−6

−120 5.09×10−6

−200 4.30×10−6

−280 3.33×10−6

−340 2.45×10−6

Regress the thermal expansion coefficient vs. temperature data to a second order polynomial.

1.00E-06

2.00E-06

3.00E-06

4.00E-06

5.00E-06

6.00E-06

7.00E-06

-400 -300 -200 -100 0 100 200

Temperature, oF

Th

erm

al e

xpan

sio

n c

oef

fici

ent,

α

(in

/in

/oF

)

Table. Data points for temperature vs

Figure. Data points for thermal expansion coefficient vs temperature.

α

46

Example 2-Polynomial Model cont.

2210 TaTaaα

We are to fit the data to the polynomial regression model

n

iii

n

iii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

T

T

a

a

a

TTT

TTT

TTn

1

2

1

1

2

1

0

1

4

1

3

1

2

1

3

1

2

1

1

2

1

The coefficients

210 , a,aa are found by differentiating the sum of thesquare of the residuals with respect to each variable and

setting thevalues equal to zero to obtain

47


The necessary summations are as follows

Temperature, T(oF)

Coefficient of thermal expansion,

α (in/in/oF)

80 6.47×10−6

40 6.24×10−6

−40 5.72×10−6

−120 5.09×10−6

−200 4.30×10−6

−280 3.33×10−6

−340 2.45×10−6

Table. Data points for temperature vs.

α 57

1

2 105580.2 i

iT

77

1

3 100472.7 i

iT

107

1

4 101363.2

i

iT

57

1

103600.3

i

i

37

1

106978.2

i

iiT

17

1

2 105013.8

i

iiT

48


1

3

5

2

1

0

1075

752

52

105013.8

106978.2

103600.3

101363.2100472.7105800.2

100472.7105800.210600.8

105800.2106000.80000.7

a

a

a

Using these summations, we can now calculate

210 , a,aa

Solving the above system of simultaneous linear equations we have

11

9

6

2

1

0

102218.1

102782.6

100217.6

a

a

a

The polynomial regression model is then

21196

2210

T101.2218T106.2782106.0217

α

TaTaa

49

Linearization of DataTo find the constants of many nonlinear models, it results in solving simultaneous nonlinear equations. For mathematical convenience, some of the data for such models can be linearized. For example, the data for an exponential model can be linearized.As shown in the previous example, many chemical and physical processes are governed by the equation,

bxaey Taking the natural log of both sides yields, bxay lnln

Let yz ln and

aa ln0

(implying)

oaea with

ba 1

We now have a linear regression model where

xaaz 10

50

Linearization of data cont.Using linear model regression methods,

_

1

_

0

1

2

1

2

11 11

xaza

xxn

zxzxna

n

i

n

iii

n

ii

n

i

n

iiii

Once 1,aao are found, the original constants of the model are found as

0

1

aea

ab

51

Example 3-Linearization of data

t(hrs) 0 1 3 5 7 9

1.000 0.891 0.708 0.562 0.447 0.355

Many patients get concerned when a test involves injection of a radioactive material. For example for scanning a gallbladder, a few drops of Technetium-99m isotope is used. Half of the technetium-99m would be gone in about 6 hours. It, however, takes about 24 hours for the radiation levels to reach what we are exposed to in day-to-day activities. Below is given the relative intensity of radiation as a function of time.

Table. Relative intensity of radiation as a function

of time

0

0.5

1

0 5 10

Rel

ativ

e in

ten

sity

of r

adia

tio

n, γ

Time t, (hours)

Figure. Data points of relative radiation intensity vs. time

52

Example 3-Linearization of data cont.

Find: a) The value of the regression

constants A an

db) The half-life of Technium-99m

c) Radiation intensity after 24 hours

The relative intensity is related to time by the equation tAe

53


tAe Exponential model given as,

tA lnln

Assuming

lnz , Aao ln and 1a we obtaintaaz

10

This is a linear relationship between

z and t

54


Using this linear relationship, we can calculate

10 , aa

n

i

n

ii

n

ii

n

i

n

iiii

ttn

ztztna

1

2

1

2

1

11 1

1

and

taza 10

where

1a

0a

eA

55

Example 3-Linearization of Data cont.

123456

013579

10.8910.7080.5620.4470.355

0.00000−0.11541−0.34531−0.57625−0.80520−1.0356

0.0000−0.11541−1.0359−2.8813−5.6364−9.3207

0.00001.00009.000025.00049.00081.000

25.000 −2.8778 −18.990 165.00

Summations for data linearization are as follows

Table. Summation data for linearization of data model

i it i iiz ln ii

zt 2it

With 6n

000.256

1

i

it

6

1

8778.2i

iz

6

1

990.18i

iizt

00.1656

1

2 i

it

56


Calculating 10 ,aa

21

2500.1656

8778.225990.186

a 11505.0

6

2511505.0

6

8778.20

a4106150.2

Since Aa ln0 0aeA

4106150.2 e 99974.0

11505.01 aalso

57


Resulting model is

te 11505.099974.0

0

0.5

1

0 5 10

Time, t (hrs)

Relative Intensity

of Radiation,

te 11505.099974.0

Figure. Relative intensity of radiation as a function of temperature using linearization of data model.

58


The regression formula is thente 11505.099974.0

b) Half life of Technetium 99 is when02

1

t

hours.t

.t.

.e

e.e.

t.

.t.

02486

50ln115050

50

9997402

1999740

115080

0115050115050

59


c) The relative intensity of radiation after 24 hours is then 2411505.099974.0 e

063200.0This implies that only

%3216.610099983.0

103200.6 2

of the radioactive

material is left after 24 hours.

60

Comparison Comparison of exponential model with and without data linearization:

With data linearization(Example 3)

Without data linearization(Example 1)

A 0.99974 0.99983

λ −0.11505 −0.11508

Half-Life (hrs) 6.0248 6.0232

Relative intensity after 24 hrs.

6.3200×10−2 6.3160×10−2

Table. Comparison for exponential model with and without data linearization.

The values are very similar so data linearization was suitable to find the constants of the nonlinear exponential model in this case.

61

62

ADEQUACY OF REGRESSION MODELS

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

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 5.86

60 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)(

2. Find the standard error of estimate

Standard error of estimate

2/

n

Ss 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)(


25283.0rS

2/

n

Ss r

T

26

25283.0

25141.0


T

ii

s

Taa

/

10 ResidualScaled

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

3

4

5

6

7

8

T

Scaled Residuals

T

ii

s

Taa

/

10 ResidualScaled

Estimateof Error Standard

Residual ResidualScaled

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

Scaled Residuals

Ti αi ResidualScaled

Residual

-340-260-180-100-2060

2.453.584.525.285.866.36

-0.285710.0685710.232860.217140.021429-0.25429

-1.13640.272750.926220.863690.085235-1.0115

25141.0/ Ts

3. Find the coefficient of determination

Coefficient of determination

n

iiir TaaS

1

210

n

iitS

1

2

t

rt

S

SSr

2

Sum of square of residuals between data and mean

n

iitS

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 TaaS

1

210

11, yx

33 , yx

22 , yx

),( nn yx

ii yx ,

iii xaayE 10

y

x

Limits of Coefficient of Determination

t

rt

S

SSr

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

iTi i

783.10tS

6750.4

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

S

SSr

2

783.10

25283.0783.10

97655.0

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

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 temperature

T α

60 6.36

40 6.24

20 6.12

0 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

Res

idua

l

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

122122

2

)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

Residu

al

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

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 =

a0+

a 1*x+

a 2*x2 +

....

.+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 =

a0+

a 1*x+

a 2*x2 +

....

.+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)]

Effect of an Outlier

Effect of Outlier

y = 2x

R2 = 1

0

5

10

15

20

25

0 2 4 6 8 10 12

Effect of Outlier

y = 3.2727x - 5.0909

R2 = 0.6879

-10

0

10

20

30

40

50

60

0 2 4 6 8 10 12

regresi 1. 2 regresi adalah diberikan sejumlah n buah data yg dimodelkan oleh persamaan model yg...

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