# Repeated measures ANOVA and Two-Factor (Factorial) ANOVA by abo20752

VIEWS: 293 PAGES: 32

• pg 1
```									   Repeated measures ANOVA and
Two-Factor (Factorial) ANOVA

A. Repeated measures: All participants experience
all of the k levels of the independent variable.
Compare to the t-test for paired samples
B. Factorial ANOVA: Treatment combinations are
applied to different participants
Compare to independent-samples t- test and one-
way ANOVA
Repeated Measures ANOVA

   Here, we partition the within sum of squares
and the within degrees of freedom.
   In a repeated measures design, differences
between treatment conditions cannot be due
to individual differences, so we subtract the
variance due to participants from the within
sum of squares, leaving us with a smaller
error term and, as with the paired samples t-
test, more power.
A repeated-measures version of the
dating study
Number of dates
Participant Soph Jr Sr Person total
Shane         2    4 6   12
Eric          1    4 8   13
Ryan          0    3 9    12
Zachary       4    1 2     7
Mathias       3     5 6   14
Totals       10 17 31 58
The F –ratio in a repeated measures
design
As always, the F – ratio compares the
variance due to treatments + error to
the variance due to error.
Therefore, we will compute SS for the
total set of scores (SSTot), within groups
(SSW), and between treatments (SSB).
Partitioning or analyzing the within
sum of squares
SSW = SSBetweenSubj + SSError

And SSBetweenSubj = S(P2/ k)- (SX)2 / N

Then, subtract to find SSError:

SSError = SSW - SSBetweenSubj
The repeated-measures ANOVA
summary table
Source      SS df MS or s2 F   p
Between Treatments
Within
Between subjects
Error
Total
Post hoc tests with repeated-
measures ANOVA
Use Tukey’s HSD or Scheffe’s test, but
with MSerror and dferror rather than
MSwithin and dfwithin.
Two-way factorial ANOVA

Partitioning the between-groups
Sum of Squares
The interaction Sum of Squares
The ANOVA summary table

Source           SS df MS F      p
Between
Within
Between participants/subjects
Error
Total
Partitioning the between-groups
Sum of Squares
   Cell notation: Rows, columns, and
interactions
   Factorial design: Fully crossed
   Set up the data so that the groups of
one variable form rows and the groups
of the other variable form columns.
Setting up the data

COLUMN_Variable
1    2    3_
| 1 | R1C1 R1C2 R1C3
ROW |
Variable| 2 | R2C1 R2C2 R2C3
An example

Number of dates/person this semester:
COLUMN___
1(So)    2(Jr)   3(Sr)_
1       7 49      2 4      9 81
(Men)     6 36      3 9    11 121
7 49      0 0    10 100
ROW            20 134    5 13    30 302
4 16     12 144 5 25
2       2 4      14 196 6 36
(Women) 1 1         15 225 7 49
7 21     41 565   18 110
The factorial ANOVA table

Source      SS df MS or s2 F   p
Between cells (Treatment)
Row (A)
Column (B)
R x C (A x B)
Within
Total
SStotal

 Calculate SStotal the same way as for the
one-way ANOVA:
SStotal = SX2 - (SX)2 / N = 1145 - 1212/ 18
= 1145 - 14641/18 = 1145 - 813.389
= 331.611
 Total df = N - 1 = 18 - 1 = 17
SSw

SSw is also computed the same as it was
for the one-way ANOVA, this time
computing SS for each R x C cell and
SSR1C1= 134 - 202 / 3 = 134 - 400/3 =0.667
SSR1C2= 13 - 52 / 3 = 13 - 25/3 = 4.667
SSR1C3= 302 - 302 / 3 = 302 - 900/3 = 2.000
SSw...

SSR2C1= 21 - 72 / 3 = 21 - 49/3 = 4.667
SSR2C2= 565 - 412 /3 = 565 - 1681/3 =4.667
SSR2C3=110 - 182 / 3 = 110 - 324/3 = 2.000
SSW= 0.667 + 4.667 + 2.000 + 4.667 +
4.667 + 2.000 = 18.668
 Within df = N - k = 18 - 6 = 12
The factorial ANOVA table

Source        SS     df MS or s2 F   p
Betweencells
Row
Column
RxC
Within        18.668 12
Total        331.611 17
SS between cells

Compute SSbetween cells the same way you
computed SSbetween in the one-way
ANOVA:
SSbetween cells= S[(SXcell)2/ncell] - (SXtotal)2/ N
= 202 + 52 + 302 + 72 + 412 + 182 - 1212/18
3      3      3    3      3        3
= 400+25+900+49+1681+324 - 813.389
3
SS between cells

= 3379 / 3 - 813.389 = 1126.333-813.389
= 312.944
   Between cells df = k - 1 = 6 - 1 = 5
The factorial ANOVA table

 Source             SS           df MS or s2* F   p
Betweencells 312.944 5
Row
Column
RxC
Within             18.668 12
Total            331.611 17
*SPSS and everyone else in the world uses MS.
SS rows

   Compute SSrows in the same way as
SSBetween, using the rows as the only
groups (pretend there are no columns):
SSrows= S[(SXrow)2/nrow] - (SXtotal)2/ N
= 552 + 662 - 813.389
9      9
= 3025 + 4356 - 813.389 = 6.722
9
SS columns

Similarly, find SScolumns using the SSBetween
formula, using columns as the only
groups:
SScolumns= S[(SXcolumns)2/ncolumns] - (SXtotal)2/
N
= 272 + 462 + 482 - 813.389
6      6     6
= 729 + 2116 + 2304 - 813.389 = 44.778
6
SS row by column interaction

 Compute the SSR x C interaction by
subtracting both the SSRows and the
SScolumns from the SSBetween cells:
SSR x C = SSBetween cells - SSRows - SSColumns
= 312.944 - 6.722 - 44.778 = 261.444
dfRows = r - 1 (number of rows - 1) = 2-1=1
dfColumns = c - 1 (number of columns - 1)= 2
dfR x C = (r - 1)(c - 1) = (1)(2) = 2
The factorial ANOVA table

Source          SS     df MS or s2 F   p
Betweencells   312.944 5
Row             6.722 1
Column         44.778 2
RxC           261.444 2
Within          18.668 12
Total          331.611 17
Computing MS or      sW2

Divide each SS by its df:
MSRows = SSRows / dfRows =6.722 / 1 = 6.722
MSCols = SSCols / dfCols = 44.778 / 2 = 22.389
MSR x C= SSRxC / dfRxC = 261.444/2 =
130.722
MSW = SSW / dfW = 18.668 / 12 = 1.556
The factorial ANOVA table

Source          SS     df MS or s2 F   p
Betweencells   312.944 5
Row             6.722 1     6.722
Column         44.778 2 22.389
RxC           261.444 2 130.722
Within          18.668 12    1.556
Total          331.611 17
F ratios

To compute F ratios, divide each
MSBetween by MSW:
FRows = MSRows / MSW = 6.722 / 1.556 = 4.32
FCols = MSCols / MSW = 22.389 / 1.556=14.39
FRxC = MSRxC / MSW = 130.722/1.556=84.01
The factorial ANOVA table

Source          SS     df MS or s2 F    p
Betweencells   312.944 5
Row             6.722 1     6.722 4.32 >.05
Column         44.778 2 22.389 14.39 <.05
RxC           261.444 2 130.722 84.01 <.05
Within          18.668 12    1.556
Total          331.611 17
Interpretation of main effects

   The main effect for rows (gender) was
not significant. We retain the null
hypothesis; the difference is due to
chance.
   The main effect for columns (class) was
significant. We reject the null
hypothesis; at least one difference is not
due to chance. Post hoc comparisons
are needed next.
Interpretation of interaction effect

   The interaction between gender (rows)
and class (columns) was significant.
The effect of class on number of dates
is different for the two genders.
   A graph of the means shows that the
most frequent dating for men occurred
among the seniors, while for women,
the most frequent dating was among the
juniors.
Interpreting the interaction...

16                             The two lines are
14                              clearly not parallel,
12                              showing the
10
Men
interaction.
8
6
Women      When there is a
4
significant
2                              interaction, interpret
0                              the main effects
So   Jr   Sr               cautiously.
Group comparisons

   Main effect
comparisons
   Interaction
comparisons
– By row variable
– By column variable

```
To top