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And, finally, we can organize all of the above into a table, along with the appropriate '''F,,CRIT,,''' values (looked up in a table like <a href="http://www.medcalc.org/manual/f-table.php">this one</a>) that we'll use for comparison and interpretation of our computations: |
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|||||||||||||| ANOVA TABLE || || ''Source'' || ''SS'' || ''df'' || ''s^2^'' || ''F,,obt,,'' || ''F,,crit,,'' || ''p'' || || rows || 63.0125 || 1 || 63.0125 || 41.06721 || 4.49 || ''p'' < 0.05 || || columns || 7.8125 || 1 || 7.8125 || 5.09165 || 4.49 || ''p'' < 0.05 || || r * c || 10.5125 || 1 || 10.5125 || 6.851324 || 4.49 || ''p'' < 0.05 || || within || 24.55 || 16 || 1.534375 || -- || -- || -- || || total || 187.225 || 19 || -- || -- || -- || -- || |
When considering the relationship among three or more variables, an interaction may arise. Interactions describe a situation in which the simultaneous influence of two variables on a third is not additive. Most commonly, interactions are considered in the context of regression analyses, but they may also be evaluated using two-way ANOVA.
A simple setting in which interactions can arise is a two-factor experiment analyzed using Analysis of Variance (ANOVA). Suppose we are interested in studying the effects of cocaine on sleep. We might design an experiment to simultaneously test whether both the use of cocaine and the duration of usage affect the number of hours a squirrel will sleep in a night. We might give half of the squirrels we test cocaine, and the other half a placebo substance (the substance variable). And we might vary the duration of usage by administering cocaine or placebo for one of two possible durations before test, 4 weeks or 12 weeks (the duration variable). We can then consider the average treatment response (e.g. number of hours slept) for each squirrel, as a function of the treatment combination that was administered (e.g. substance and duration). The following table shows one possible situation:
4-Week Placebo (Control) |
4-Week Cocaine |
12-Week Placebo (Control) |
12-Week Cocaine |
7.5 |
5.5 |
8.0 |
5.0 |
8.0 |
3.5 |
10.0 |
4.5 |
6.0 |
4.5 |
13.0 |
4.0 |
7.0 |
6.0 |
9.0 |
6.0 |
6.5 |
5.0 |
8.5 |
4.0 |
There are three null hypotheses to be tested:
H01: Both substance groups sleep for the same number of hours on average.
H02: Both treatment duration groups sleep for the same number of hours on average.
H03: The two factors are independent or there is no interaction effect.
We can start by computing the group means (for each cell, row, and column):
Group Means |
|||
|
4-Week |
12-Week |
All Durations |
Placebo |
7 |
9.7 |
8.35 |
Cocaine |
4.9 |
4.7 |
4.8 |
All Substances |
5.95 |
7.2 |
6.575 |
And by calculating the sum of squares (SS) for each group (cell):
4-Week Placebo Dataset = {7 8 6 7 6.5}, M = 7
SS4-Wk Placebo = (7-7)2 + (8-7)2 + (6-7)2 + (7-7)2 + (6.5-7)2 = 2.25
4-Week Cocaine = {5.5 3.5 4.5 6 5}, M = 4.9
SS4-Wk Cocaine = (5.5-4.9)2 + (3.5-4.9)2 + (4.5-4.9)2 + (6-4.9)2 + (5-4.9)2 = 3.7
12-Week Placebo = {8 10 13 9 8.5}, M = 9.7
SS12-Wk Placebo = (8-9.7)2 + (10-9.7)2 + (13-9.7)2 + (9-9.7)2 + (8.5-9.7)2 = 15.8
12-Week Cocaine = {5 4.5 4 6 4}, M = 4.7
SS12-Wk Cocaine = (5-4.7)2 + (4.5-4.7)2 + (4-4.7)2 + (6-4.7)2 + (4-4.7)2 = 2.8
Now, we'll calculate the SSB:
SSB = n [( M4-Wk Placebo - MGroup )2 + ( M4-Wk Cocaine - MGroup)2 + ( M12-Wk Placebo - MGroup )2 + ( M12-Wk Cocaine - MGroup)2]
= 5 [(7 - 6.575 )2 + (4.9 - 6.575)2 + (9.7 - 6.575)2 + (4.7 - 6.575)2]
= 5 [0.180625 + 2.805625 + 9.765625 + 3.515625]
= 5 [16.2675]
= 81.3375
Now, we'll calculate the SSW:
SSW = SS4-Wk Placebo + SS4-Wk Cocaine + SS12-Wk Placebo + SS12-Wk Cocaine
= 2.25 + 3.7 + 15.8 + 2.8
= 24.55
dfW = N - rc
= 20 - (2 * 2)
= 16
sW2 = SSW / dfW
= 24.55 / 16
= 1.534375
Now, we'll calculate the SSR:
SSR = n [( MPlacebo - MGroup )2 + ( M Cocaine - MGroup)2]
= 10 [(8.35 - 6.575)2 + (4.8 - 6.575)2]
= 10 [3.150625 + 3.150625]
= 10 [6.30125]
= 63.0125
dfR = r - 1
= 2-1
= 1
sR2 = SSR / dfR
= 63.0125 / 1
= 63.0125
Now, we'll calculate the SSC:
SSC = n [( M4-Week - MGroup )2 + ( M 12-Week - MGroup)2]
= 10 [(5.95 - 6.575)2 + (7.2 - 6.575)2]
= 10 [0.390625 + 0.390625]
= 10 [0.78125]
= 7.8125
dfC = c - 1
= 2-1
= 1
sC2 = SSC / dfC
= 7.8125 / 1
= 7.8125
Now, we'll calculate the SSRC:
SSRC = SSB - SSR - SSC
= 81.3375 - 63.0125 - 7.8125
= 10.5125
dfRC = (r - 1)(c - 1)
= (2-1)(2-1)
= 1
sRC2 = SSRC / dfRC
= 10.5125 / 1
= 10.5125
Now, we'll calculate the SST:
SST = SSB + SSW,'' + ''SSR'' + ''SSC''+ ''SSRC,,
= 81.3375+ 24.55 + 63.0125 + 7.8125 + 10.5125
= 187.225
dfT = N - 1
= 20-1
= 19
Now, we'll calculate the F values:
FR = sR2 / sW2
= 63.0125 / 1.534375
= 41.06721
FC = sC2 / sW2
= 7.8125 / 1.534375
= 5.09165
FRC = sRC2 / sW2
= 10.5125 / 1.534375
= 6.851324
And, finally, we can organize all of the above into a table, along with the appropriate FCRIT values (looked up in a table like <a href="http://www.medcalc.org/manual/f-table.php">this one</a>) that we'll use for comparison and interpretation of our computations:
FCRIT (1, 16) α=0.5 = 4.49
ANOVA TABLE |
||||||
Source |
SS |
df |
s2 |
Fobt |
Fcrit |
p |
rows |
63.0125 |
1 |
63.0125 |
41.06721 |
4.49 |
p < 0.05 |
columns |
7.8125 |
1 |
7.8125 |
5.09165 |
4.49 |
p < 0.05 |
r * c |
10.5125 |
1 |
10.5125 |
6.851324 |
4.49 |
p < 0.05 |
within |
24.55 |
16 |
1.534375 |
-- |
-- |
-- |
total |
187.225 |
19 |
-- |
-- |
-- |
-- |