Split-plot_anova

R e s e a r c h p a p e r
Effects of alcohol and caffeine on driving ability
4 conditions:
S p l i t - p l o t A N O V A
– No alcohol; no caffeine
– alcohol; no caffeine
– No alcohol; caffeine
– Alcohol; caffeine
D r i v i n g i n s i m u l a t o r
Error rate
Split-plot design
S p l i t - p l o t d e s i g n
S p l i t - p l o t d e s i g n
N o Alcohol
A l c o h o l a s b e t w e e n - p a r t i c i p a n t s f a c t o r
Participant
No caffeine
Caffeine Participant
No caffeine
Caffeine
C a f f e i n e a s w i t h i n - p a r t i c i p a n t s f a c t o r
M e a n i n g :
– 12 participants in the no alcohol condition
– 12 participants in the alcohol condition
– But all of them in the caffeine/no caffeine conditions
S P S S o u t p u t : ” b e t w e e n e f f e c t s ”
S P S S o u t p u t : “ w i t h i n e f f e c t s ”
Main effect of caffeine
Main effect of alcohol
Tests of Within-Subjects Effects
Tests of Between-Subjects Effects
Interaction between
caffeine and alcohol
S P S S o u t p u t : c e l l p l o t
R e s e a r c h p a p e r
Results:
T h e n u m b e r o f d r i v i n g e r r o r s w a s a n a l y z e d w i t h a
s p l i t - p l o t A N O V A w i t h a l c o h o l a s t h e b e t w e e n -
p a r t i c i p a n t s f a c t o r a n d c a f f e i n e a s t h e w i t h i n -
participants factor. The test indicated a main
effect of alcohol (F(1, 22) = 382.28, p < 0.001) and
of caffeine (F(1, 22) = 521.56, p < 0.001). The
i n t e r a c t i o n b e t w e e n a l c o h o l a n d c a f f e i n e w a s
s i g n i f i c a n t a s w e l l ( F ( 1 , 2 2 ) = 5 7 . 9 5 , p < 0 . 0 0 1 ) .
C o m p a r i s o n
1 . E x a m p l e
B e t w e e n - p a r t i c i p a n t s d e s i g n :
– Alcohol: (F(1, 44) = 339.8, p < 0.001)
– Caffeine: (F(1,44) = 515.4, p < 0.001)
Do boys and girls differ in the ability to perceive
colours?
– Alcohol x caffeine: F(1,44) = 37.8, p < 0.001.
W i t h i n - p a r t i c i p a n t s d e s i g n :
T h e s t u d y a s s u m e d t h a t g i r l s w i l l b e b e t t e r t h a n
boys at perceiving differences in colours f r o m a
– Alcohol (F(1, 11) = 577.9, p < 0.001)
– Caffeine (F(1, 11) = 692.5, p < 0.001).
v e r y e a r l y a g e . T h e y t h e r e f o r e t e s t e d t w o d i f f e r e n t
age groups (5-year-olds and 11-year-olds) on a
– Alcohol x Caffeine: (F(1, 11) = 52.8, p < 0.001).
s t a n d a r d c o l o u r p e r c e p t i o n t e s t a n d c o m p a r e d
M ix e d d e si g n :
t h e p e r f o r m a n c e ( m a r k e d o u t o f 1 0 ) o f b o y s a n d
– Alcohol (F(1, 22) = 382.28, p < 0.001)
– Caffeine (F(1, 22) = 521.56, p < 0.001).
– Alcohol x Caffeine: (F(1, 22) = 55.25, p < 0.001).
P e r f o r m a n c e
C e l l p l o t
5 - y e a r - o l d s
1 1 - y e a r - o l d s
2 . E x a m p l e
A g e : F ( 1 , 4 4 ) = 1 0 . 7 2 ; p = 0 . 0 0 2
Has the academic ability fallen in the last 20
years?
G e n d e r : F ( 1 , 4 4 ) = 4 8 . 8 6 2 ; p < 0 . 0 0 1
T h e s t u d y c o m p a r e d t h e A - l e v e l p e r f o r m a n c e o f a
s a m p l e o f s t u d e n t s w h o t h e e x a m s i n 1 9 9 7 a n d a
A g e x G e n d e r : F ( 1 , 4 4 ) = 2 2 . 6 9 ; p = 0 . 0 0 6
s a m p l e w h o t o o k t h e m i n 1 9 7 7 . E a c h h a d t a k e n
a n e x a m i n a t i o n i n b o t h E n g l i s h a n d M a t h e m a t i c s .
In order to ensure that the exams are marked to
t h e s a m e c r i t e r i a t h e s a m p l e s a r e r e - m a r k e d b y
e x a m i n e r s .
N e w m a r k s
C e l l p l o t
S t u d e n t s f r o m 1 9 7 7
S t u d e n t s f r o m 1 9 9 7
M a t h e m a t i c s
M a t h e m a t i c s
M o r e i n d e p e n d e n t v a r i a b l e s
S u b j e c t : F ( 1 , 2 2 ) = 0 . 0 0 1 ; p = 0 . 9 8 2
– 2 x 2 factorial design
– 3 x 3 factorial design
Y e a r : F ( 1 , 2 2 ) = 5 . 8 2 8 ; p = 0 . 0 2 5
P o s s i b l e : a r b i t r a r y n u m b e r o f i n d e p e n d e n t
variables and levels
E x a m p l e s :
– 3 x 4 x 5 factorial design (Three-way ANOVA)
Subject x Year: F(1,22) = 0.088; p = 0.982
– 4 x 4 x 2 x 2 x 2 x 5 x 6 factorial design
H o w e v e r , m o r e t h e n 3 i n d e p e n d e n t v a r i a b l e s
d o e s n o t m a k e s e n s e !
E x a m p l e f o r 3 x 2 x 2 d e s i g n
T y p i c a l V i s u a l S e a r c h r e s u l t s
Does advance information help in visual search?
Reaction time
N u m b e r o f i t e m s
E x p e r i m e n t a l p r o c e d u r e
Target: L, Chevron, absent (catch-trials)
A d v a n c e I n f o r m a t i o n ( p r i m e ) : V a l i d , I n v a l i d ,
N u m b e r o f i t e m s : 4 , 6
2 x 3 x 2 A N O V A r e p e a t e d - m e a s u r e s
Validity: F(2,34)=10.624,p < 0.001
T a r g e t : F ( 1 , 1 7 ) = 4 7 . 4 7 7 , p < 0 . 0 0 1
I t e m s : F ( 1 , 1 7 ) = 6 0 . 3 0 6 , p < 0 . 0 0 1
Validity x target: F(2, 34) = 19.515, p < 0.001
V a l i d i t y x i t e m s : F ( 2 , 3 4 ) = 0 . 3 7 1 , p = 0 . 6 9 3
Mean Reaction Time (ms)
T a r g e t x i t e m s : F ( 1 , 1 7 ) = 1 2 . 2 0 5 , p = 0 . 0 0 3
N u m b e r o f I t e m s
V a l i d i t y x t a r g e t x i t e m s : F ( 2 , 3 4 ) = 6 . 1 1 6 , p = 0 . 0 0 5
R e s u l t s : S i m p l e e f f e c t
P o w e r o f t e s t
P r o b a b i l i t y o f c o r r e c t l y r e j e c t i n g a f a l s e H 0
T a r g e t C h r e v o n:
– Validity : F(2,34)=17.885, p < 0.001
O r n o t m a k i n g T y p e I I e r r o r
– Item: F(1,17)=23.638, p < 0.001
– Validity x items: F(2,34)=4.629,p = 0.017
T a r g e t L :
Decision
State of the world:
State of the world:
– Validity: F(2,34)=1.752, p=0.189
– Item: F(1,17)=54.152, p < 0.000
– Validity x items: F(2,34)=2.427, p = 0.103
Type I error
Correct decision
Fail to reject H
Correct decision
Type II error
I n f l u e n c e s o n p o w e r o f t e s t
§ α-level (Probability of Type I error)
Grand mean
T r u e d i f f e r e n c e b e t w e e n t h e n u l l h y p o t h e s i s a n d
the alternative hypothesis
S a m p l e s i z e
Variance
P r o p e r t i e s o f t h e t e s t e m p l o y e d
Grand mean

Source: http://www.dietmar-heinke.co.uk/files/split-plot_anova1.pdf