Lecture 10. Factorial experiments (2-way ANOVA etc)

25 Pages·2014·0.43 MB·English

Lecture 10. Factorial experiments
(2-way ANOVA etc)
Jesper Ryd´en
Matematiska institutionen, Uppsala universitet
[email protected]
Regression and Analysis of Variance • autumn 2014
A factorial experiment
Two factors A and B, each run at two levels (“low” and “high”)
Two-factor factorial experiment Interaction plot (no interaction)
(responses in corners)
Another factorial experiment
Two factors A and B, each run at two levels (“low” and “high”)
Two-factor factorial experiment Interaction plot (interaction)
(responses in corners)
Data arrays (Battery data)
Response: Life (in hours) of battery.
Factor A: Material type (1, 2 or 3).
Factor B: Temperature (in ◦F), levels 15, 70 or 125.
This is a 32 factorial design.
Temperature can be controlled in the laboratory.
The effects model
The effects model for a two-factor factorial design:

i = 1,2,...,a

y = µ+τ +β +(τβ) +(cid:15) j = 1,2,...,b
ijk i j ij ijk
 k = 1,2,...,n
µ Overall mean effect
τ Effect of the ith level of the row factor A
i
β Effect of the jth level of the column factor B
j
(τβ) Effect of the interaction between τ and β
ij i j
(cid:15) Random error component
ijk
Both factors are assumed to be fixed and
a n
(cid:88) (cid:88)
τ = 0, β = 0
i j
i=1 j=1
and for interactions
a b
(cid:88) (cid:88)
(τβ) = (τβ) = 0.
ij ij
i=1 j=1
The means model
The means model for a two-factor factorial design:

i = 1,2,...,a

y = µ +(cid:15) j = 1,2,...,b
ijk ij ijk
 k = 1,2,...,n
where the mean of the ijth cell is
µ = µ+τ +β +(τβ) .
ij i j ij
Testing hypotheses
Testing equality of row treatment effects:
H : τ = τ = ··· = τ = 0
0 1 2 a
H : at least one τ (cid:54)= 0
1 i
Testing equality of column treatment effects:
H : β = β = ··· = β = 0
0 1 2 b
H : at least one β (cid:54)= 0
1 j
Test of interaction:
H : (τβ) = 0, for all i,j
0 ij
H : at least one (τβ) (cid:54)= 0
1 ij
Estimation of parameters
The Least-Squares method applied on the effects model with
constraints result in the point estimates
µ = y¯ ,
(cid:98) •••
τ = y¯ −y¯ ,
(cid:98)i i•• •••
β(cid:98)j = y¯•j•−y¯•••,
(cid:26)
i = 1,2,...,a
((cid:100)τβ) = y¯ −y¯ −y¯ +y¯ ,
ij ij• i•• •j• ••• j = 1,2,...,b
Fitted value:
y(cid:98)ijk = µ(cid:98)+τ(cid:98)i +β(cid:98)j +((cid:100)τβ)ij = y¯ij•.
ANOVA table: Fixed-effects case, two factors
Source of Sum of Degrees of
Variation Squares Freedom Mean Square F
0
A treatments SS a−1 MS = SSA F = MSA
A A a−1 0 MSE
B treatments SS b−1 MS = SSB F = MSB
B B b−1 0 MSE
Interaction SS (a−1)(b−1) MS = SSAB F = MSAB
AB AB (a−1)(b−1) 0 MSE
Error SS ab(n−1) MS = SSE
E E ab(n−1)
Total SS abn−1
T
SS = SS + SS
T Treat E
(cid:124)(cid:123)(cid:122)(cid:125) (cid:124) (cid:123)(cid:122) (cid:125) (cid:124)(cid:123)(cid:122)(cid:125)
N−1 ab−1 N−ab
= SS +SS + SS + SS
A B AB E
(cid:124)(cid:123)(cid:122)(cid:125) (cid:124)(cid:123)(cid:122)(cid:125) (cid:124)(cid:123)(cid:122)(cid:125) (cid:124)(cid:123)(cid:122)(cid:125)
a−1 b−1 (a−1)(b−1) N−ab
So-called machine formulae for computation of sums of squares,
see the textbook.
ANOVA table (battery data)