PAIRWISE t-TEST AND
ANALYSIS OF VARIANCE
Emese Vágó, Sándor Kemény
Budapest University of Technology and Economics
Outline
I. Summary of paired t-test and analysis of variance
II. Similarities and differences in the two procedures
III. Comparison of the models
IV. Comment
IV. Conclusions
Example
1
2
3
After
diet
65
45
78
Before
diet
70
49
86
…
…
…
People
The effect of a diet on weight was investigated using ten people.
Results:
10
79
85
1
2
3
y 2n
H 0 : E (d )  0
n
d
d
…
y 1n
…
n
G ro u p
d i= y 1 i- y 2 i
1
2
y 11 y 21
d1
y 12 y 22
d2
y 13 y 23
d3
…
…
Pairs
Paired t-test
i 1
n
i
n
s 
2
d
dn
d  E (d )

sd / n
d (n  1)n
n
 (d
i 1
i
 d)
~ t n -1
 (d
i 1
i
 d)
n 1
Analysis of variance
1
2
3
2
y 11
y 12
y 13
y 21
y 22
y 23
…
n
1
…
…
Random effect
(B)
F ix e d e ffe c t (A )
y 1n
y 2n
yij: weight of jth men
before(i=1)/after(i=2) diet
r: levels of A (2)
n: levels of B (10)
r
MS A

MS AB
 y
i 1
 y
r
n
i 1 j 1
 y.. 
2
i.
 yi.  y. j  y.. 
2
ij
( n  1)n ~ F1,n-1
Similarities and differences
t
2
n 1
 F1,n1
?
Conditions of the two analysis differ in
spite of the same statistics. Why is it so?
?
Same statistic
Same model
Assumptions - paired t-test

d j ~ N d , 
2
d

       1  2
2
d
2
1
2
2
Product moment correlation,
can not be estimated
 
~ N  ,  
y1 j ~ N 1 , 
y2 j
2
2
1
2
2
y1 j  1  1 j
1 j ~ N (0,  )
y2 j   2   2 j
 2 j ~ N (0, )
2
1
Assumptions of independence for d j , y1 j , y2 j
2
2
Assumptions - analysis of variance


yij     i   j  ij   ij
 ij ~ N (0,  e2 )
 j ~ N (0,  b2 )
2
yij ~ N (  i , b2   ab
  e2 )

Assumptions of independence for
- yij
- levels of B
2
ij ~ N (0,  ab
)
Comparison of assumptions
Paired t-test
Analysis of variance
yij ~ N (i ,  )
yij ~ N (  i ,     )
2
b
2
i
2
ab
i     i
If
If
  
2
1
2
2
 12   22
2
2
 2   b2   ab
  e2
the models do not
correspond
2
e
Comparison of assumptions
ANOVA method may be used with weaker assumptions
about homogeneity of variances for:
- one random, and
- one fixed effect (number of levels is two) design
- sample sizes are one
- testing only the effect of fixed factor
Comment - distribution of the random effect

d ~ N  d ,  d2

yij ~ N 1 , 
2
i


yij ~ ?
 d2   12   22   1  2
y1 j  1  1 j
y2 j   2   2 j
1   j  1 j  y1 j
 2   j   2 j  y2 j
 ij ~ N (0,  i2 )
 ij ~ N (0,  i2 )
d j  1   2  1 j   2 j
Comment - sample sizes larger then one
Sample size: 1
Sample size: q
y1 jk  1   j  1 jk
y2 jk   2   j   2 jk
1   j  1 j  y1 j
 2   j   2 j  y2 j
y1 j  1   j  1 j .
y2 j .   2   j   2 j .
 ij ~ N (0,  i )
 ij . ~ N (0,  i2 / q)
d j  1   2  1 j   2 j
d j  1   2  1 j .   2 j .
2
Consequence
ANOVA method may be used with weaker assumptions
about homogeneity of variances for testing the effect of a
2-level fixed factor, if the other factor is
- random, or fixed and
- sample sizes are one, or larger then one.
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