experimental design and statistical analyses of data lesson 1: general linear models and design of...

Experimental design and statistical analyses of data

Lesson 1:

General linear models and design of experiments

Examples of General Linear Models (GLM)

Simple linear regression:

xy 10

Ex: Depth at which a white disc is no longer visible in a lake

y = depth at disappearancex = nitrogen concentration of water

0 2 4 6 8 10

N/volume water

0

2

4

6

8

10

Dep

th (

m)

The residual ε expressesthe deviation between the model and the actual observation

β0

Intercept

β1

Slope

Dependentvariable

Independentvariable

Polynomial regression:

Ex:: y = depth at disappearancex = nitrogen concentration of water

0 2 4 6 8 10

N/volume water

0

2

4

6

8

10

Dep

th (

m)

2210 xxy

Multiple regression:

21322110 xxxxy

Eks: y = depth at disappearancex1 = Concentration of N

x2 = Concentration of P

02

4

6

8Concentration of N

0

2

4

6

8

Concentration of P

0

2

4

6

8

10

Depth

0

2

4

6

8

10

Depth

Ex: y = depth at disappearancex1 = Blue disc

x2 = Green disc

White Blue Green

Disc color

0

2

4

6

8

10

De

pth

22110 xxy

x1= 0; x2 = 0x1= 1; x2= 0x1= 0; x2= 1

Analysis of variance (ANOVA)

Analysis of covariance (ANCOVA):

3253143322110 xxxxxxxy

Ex: y = depth at disappearancex1 = Blue disc

x2 = Green disc

x3 = Concentration of N

0 2 4 6 8 10

Concentration of N

0

2

4

6

8

10

Dep

th

Nested analysis of variance:

jiiy )(

Ex: y = depth at disappearanceαi = effect of the ith lake

β(i)j = effect of the jth measurement in the ith lake

What is not a general linear model?

y = β0(1+β1x)

y = β0+cos(β1+β2x)

Other topics covered by this course:

• Multivariate analysis of variance (MANOVA)

• Repeated measurements

• Logistic regression

Experimental designs

Examples

Randomised design

• Effects of p treatments (e.g. drugs) are compared

• Total number of experimental units (persons) is n

• Treatment i is administrated to ni units

• Allocation of treatments among units is random

Example of randomized design

• 4 drugs (called A, B, C, and D) are tested (i.e. p = 4)

• 12 persons are available (i.e. n = 12)

• Each treatment is given to 3 persons (i.e. ni

= 3 for i = 1,2,..,p) (i.e. design is balanced)

• Persons are allocated randomly among treatments

n

yy ij

Drugs

A B C D Total

y1A

y2A

y3A

y1B

y2B

y3B

y1C

y2C

y3C

y1D

y2D

y3D

A

jAA n

yy

B

jBB n

yy

C

jCC n

yy

D

jDD n

yy

DD

CC

BB

AA

yy

yy

yy

yy

Note!Different persons

DD

CC

BB

AA

yy

yy

yy

yy

0Ay

10 By AB yy 1

ADD

ACC

yyy

yyy

330

220

30

20

10

0

11 x

12 x

13 x

3322110 xxxy

Source Degrees of freedom

Estimate of

Treatments ( )

Residuals

1

p - 1 = 3

n-p = 8

Total n = 12

0

321

Randomized block design

• All treatments are allocated to the same experimental units

• Treatments are allocated at random

B C B

A B D

D A A

C D C

Blocks (b = 3)

Treatments (p = 4)

Treatments

Persons

A B C D Average

1

2

3

Average

Cy1 Dy1

Ay2

Ay3

Cy2By2 Dy2

By3 Cy3 Dy3

Ay By

1y

2y

3y

Cy Dy y

55443322110 xxxxxy

Blocks (b-1) Treatments (p-1)

Ay1 By1


Estimate of

Blocks (persons)

Treatments ( drugs )

Residuals

1

b - 1 = 2

p-1 = 3

n-[(b-1)+(p-1)+1] = 6

Total n = 12

0

Randomized block design

Double block design (latin-square)Person

Sequence

1 2 3 4

1 B D A C

2 A C D B

3 C A B D

4 D B C A

Rows (a = 4)

Columns (b = 4)

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Sequence (a-1) Persons (b-1)

Drugs (p-1)


Estimate of

Rows (sequences)

Blocks (persons)

Treatments ( drugs )

Residuals

1

a-1 = 3

b - 1 = 3

p-1 = 3

n-[3(p-1)+1] = 6

Total n = p2 = 16

0

Latin-square design

Factorial designs

• Are used when the combined effects of two or more factors are investigated concurrently.

• As an example, assume that factor A is a drug and factor B is the way the drug is administrated

• Factor A occurs in three different levels (called drug A1, A2 and A3)

• Factor B occurs in four different levels (called B1, B2, B3 and B4)

Factorial designs

Factor B

Factor A

B1 B2 B3 B4 Average

A1 y11 y12 y13 y14

A2 y21 y22 y23 y24

A3 y31 y32 y33 y34

Average

1y

2y

3y

1y 2y 3y 4y y

55443322110 xxxxxyij

Effect of A Effect of B No interaction between A and B

Factorial experiment with no interaction

• Survival time at 15oC and 50% RH: 17 days



• What is the expected survival time at 25oC and 80% RH?

• An increase in temperature from 15oC to 25oC at 50% RH decreases survival time by 9 days

• An increase in RH from 50% to 80% at 15oC increases survival time by 2 days

• An increase in temperature from 15oC to 25oC and an increase in RH from 50% to 80% is expected to change survival time by –9+2 = -7

days


10 15 20 25 30

Temperature (oC)

0

5

10

15

20

25S

urv

ival

tim

e (d

ays)

50 % RH

80 % RH

10 15 20 25 30

Temperature (oC)

0

5

10

15

20

25S

urv

ival

tim

e (d

ays)


22110 xxyij

0

1

2

Factorial experiment with interaction

10 15 20 25 30

Temperature (oC)

0

5

10

15

20

25S

urv

ival

tim

e (d

ays)

0

1

2

3

21322110 xxxxyij

Factorial designs

Factor B

Factor A

B1 B2 B3 B4 Average

A1 y11 y12 y13 y14

A2 y21 y22 y23 y24

A3 y31 y32 y33 y34

Average

1y

2y

3y

1y 2y 3y 4y y

Effect of A Effect of B

5211421032951841731655443322110 xxxxxxxxxxxxxxxxxyij

Interactions between A and B


Estimate of

Factor A (drug)

Factor B (administration)


Residuals

1

a-1 = 2

b - 1 = 3

(a-1)(b-1) = 6

n- ab = 0

Total n = ab = 12

0

Two-way factorial designwith interaction, but without replication


Estimate of

Factor A (drug)


Residuals

1

a-1 = 2

b - 1 = 3

n- a-b+1 = 6

Total n = ab = 12

0

Two-way factorial designwithout replication

Without replication it is necessary to assume no interaction between factors!


Estimate of

Factor A (drug)



Residuals

1

a-1

b - 1

(a-1)(b-1)

ab( r-1)

Total n = rab

0

Two-way factorial designwith replications


Estimate of

Factor A (drug)



Residuals

1

a-1 = 2

b – 1 = 3

(a-1)(b-1) = 6

ab( r-1) = 12

Total n = rab = 24

0

Two-way factorial designwith interaction (r = 2)

Factor CFactor B

Factor A

Factor A

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Factor B Factor C

Three-way factorial design

Factor A

42203219101189117811671156114511341123111 xxxxxxxxxxxxxxxxxxxx 10 Main effects

31 Two-way interactions

107272972718727094145841441031439314283141 xxxxxxxxxxxxxxxxxxxxxxxx

30 Three-way interactions


Estimate of

Factor A

Factor B

Factor C


Interactions between A and C

Interactions between B and C

Interactions between A, B and C

Residuals

1

a-1 = 2

b – 1 = 5

c-1 = 3

(a-1)(b-1) = 10

(a-1)(c-1) = 6

(b-1)(c-1) = 15

(a-1)(b-1)(c-1) = 30

abc( r-1) = 0

Total n = rabc = 72

0

Three-way factorial design

Why should more than two levels of a factor be used in a factorial design?

Two-levels of a factor

10 15 20 25 30

Temperature (oC)

0

5

10

15

20

25

30S

urv

ival

tim

e (d

ays)

10 15 20 25 30

Temperature (oC)

0

5

10

15

20

25

30S

urv

ival

tim

e (d

ays)

Three-levels

factor qualitative

22110 xxy

1

Low Medium High

0

2

10 15 20 25 30

Temperature (oC)

0

5

10

15

20

25

30S

urv

ival

tim

e (d

ays)

Three-levels

factor quantitative

2210 xxy

Why should not many levels of each factor be used in a factorial

design?

Because each level of each factor increases the number of

experimental units to be used

For example, a five factor experiment with four levels per factor yields 45 = 1024 different combinations

If not all combinations are applied in an experiment, the design is partially factorial

experimental design and statistical analyses of data lesson 1: general linear models and design of...

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