to Introduction Statistics in
Psychology
PSY 201
Professor Greg Francis
Lecture 36
ANalysis Of VAriance
The last topic!
REPEATED MEASURES
SUM OF SQUARES
one way ANOVA deals with
independent samples
scores for an individual are
dependent
scores for dierent individuals are
independent
we want to consider one situation
with a dependency
suppose we repeat our
measurements from the same
individuals at dierent times
e.g., tracking health patterns
across years, grades throughout
school,...
SST = SSI + SSO + SSRes
where
SST is the total sum of square
SSI is the variation among individuals
SSO is the variation among test
occasions
SSRes is any other type of variation
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INDIVIDUALS
OBSERVATIONS
RESIDUAL
the combined variation among
individuals is
the combined variation across
observations is
we need a term that correspond to
SSW
SSI = iK (X i X )2
SSO = k n (X k X )2
where
X
X i = k ik
K
is the average for the ith
individual across all observations
where
X
X k = i ik
n
is the average for the k th
observation across all subjects
SSI deviation of individual means
from overall mean
SSO deviation of observation
mean from overall mean
does not correspond to SSW or
SSB in the normal ANOVA
similar to SSB in the normal
ANOVA
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we can directly calculate the total
sum of squares
SST =
ki
(Xik X )2
if there is variation beyond SSI
and SSO , we can calculate it as
SSRes = SST SSI SSO
this is similar to SSW
factors out variation due to
individuals and variation due to
observations
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VARIANCE ESTIMATES
VARIANCE ESTIMATES
F RATIO
SSRes > 0 due to random
sampling (choice of individuals)
SSRes
M SRes =
(K 1)(n 1)
estimates the variance of the
population distribution
SSO can vary due to random
sampling, or due to dierences
across observations
as before we compare these
estimates with the F statistic
M SO
F=
M SRes
if H0 is true
H0 : 1 = 2 = ... = K
then there are no dierences across
observations, so all variation must
be due to random sampling. So,
SSO
M SO =
K 1
estimates the variance of the
population distribution if H0 is
true
the degrees of freedom associated
with this estimate is
(K 1)(n 1)
if H0 is true
F 1.0
if H0 is not true
F > 1.0
look up Fcv for (K-1), (K-1)(n-1)
degrees of freedom
everything else is the same as
before
otherwise it overestimates it
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EXAMPLE
(1) HYPOTHESIS
(2) CRITICAL VALUE
H 0 : 1 = 2 = 3
for the numerator (observation
sum of squares) we have
A school principal traces reading
comprehension scores on a
standardized test for a random
sample of dyslexic students across
three The years. data are given
below. Complete the ANOVA
using = 0.05.
Student Third Grade Fourth Grade Fifth Grade
1
2.8
3.2
4.5
2
2.6
4.0
5.1
3
3.1
4.3
5.0
4
3.8
4.9
5.7
5
2.5
3.1
4.4
6
2.4
3.1
3.9
7
3.2
3.8
4.3
8
3.0
3.6
4.4
Tk
23.40
30.00
37.30
ki
Ha : i = k for some i and k
test with = 0.05
df = K 1 = 3 1 = 2
for the denominator (residual sum
of squares) we have
df = (K 1)(n1) = (31)(81) = 14
Ti
10.50
11.70
12.40
14.40
10.00
9.40
11.30
11.00
90.70
so from the F distribution table
we nd the F critical value to be
Fcv = 3.74
2
Xik = 361.43
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(3) TEST STATISTIC
(3) TEST STATISTIC
there are computational formulas that help
manage the calculations
2
T
N
where T is the sum of all scores
SST =
k
i
2
Xik
SSO =
so
(90.70)2
= 361.43 342.77
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SST = 18.66
SST = 361.43
and
T2
T2
SSI = i
K
N
i
k
2
Tk T 2
n
N
(23.40)2 (30.0)2 (37.30)2
+
+
342.77
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SSO = 354.86 342.77 = 12.09
SSO =
so any remaining variation is residual
SSRes = SST SSI SSO
SSRes = 18.66 5.67 12.09 = 0.90
again, these cannot be negative!
(3) TEST STATISTIC
now calculate
SSO
12.09
M SO =
=
= 6.05
K 1
2
and
SSRes
0.90
M SRes =
=
= 0.06
(K 1)(n 1)
14
and get the F statistic
M SO
6.05
F=
=
= 100.83
M SRes 0.06
where Ti is the sum of observations for the ith
individual
so
SSI = 348.44 342.77 = 5.67
Note, this cannot be negative!
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(4) INTERPRET
SUMMARY TABLE
ASSUMPTIONS
since
F = 100.83 > 3.74 = Fcv
we reject H0.
there is evidence that the reading
scores for these subjects are
dierent across the years
Source of Sum of Degrees of Variance
variation
squares freedom estimate F ratio
Occasions
12.09
2
6.05
100.83
Individuals 5.67
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Residual
0.90
14
0.06
Total
18.66
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With = 0.05, Fcv = 3.74. Therefore H0 is
rejected.
ANOVA for repeated measures depends
on four assumptions
1. The sample was randomly selected for
a population.
2. The dependent variable (e.g., reading
scores) is normally distributed in the
population.
3. The population variances for the test
occasions are equal. (homogeneity of
variance)
4. The population correlation coecients
between pairs of test occasion scores
are equal.
deviation from (2) tends to not cause
serious problems
deviations from (3) and (4) can be
compensated for sometimes
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CONCLUSIONS
NEXT TIME
restrictions on use of one-way
ANOVA
nish up details of class
strength of association
ANOVA repeated measures
Study guide for nal exam
How does it feel to not be a
schmuck?
partitioning the variance
accounts for dependence among
scores for same subjects
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