Transcript One-Way ANOVA Independent Samples Basic Design • • • • Grouping variable with 2 or more levels Continuous dependent/criterion variable H: 1 = 2 = ...

One-Way ANOVA
Independent Samples
Basic Design
•
•
•
•
Grouping variable with 2 or more levels
Continuous dependent/criterion variable
H: 1 = 2 = ... = k
Assumptions
– Homogeneity of variance
– Normality in each population
The Model
• Yij =  + j + eij, or,
• Yij -  = j + eij.
• The difference between the grand mean
() and the DV score of subject number i
in group number j
• is equal to the effect of being in treatment
group number j, j,
• plus error, eij
Four Methods of Teaching ANOVA
Do these four samples differ enough from
each other to reject the null hypothesis
that type of instruction has no effect on
mean test performance?
Group
A
B
C
D
1
2
6
7
2
3
7
8
Score
2
3
7
8
2
3
7
8
3
4
8
9
Error Variance
• Use the sample data to estimate the
amount of error variance in the scores.
s  s  ....  s
MSE 
k
2
1
2
2
2
k
• This assumes that you have equal sample
sizes.
• For our data, MSE = (.5 + .5 + .5 + .5) / 4
= 0.5
Among Groups Variance
2
MSA  n  smeans
•
•
•
•
Assumes equal sample sizes
VAR(2,3,7,8) = 26 / 3
MSA = 5  26 / 3 = 43.33
If H is true, this also estimates error
variance.
• If H is false, this estimates error plus
treatment variance.
F
• F = MSA / MSE
• If H is true, expect F = error/error = 1.
• If H is false, expect
error  treatment
F
1
error
p
•
•
•
•
•
F = 43.33 / .5 = 86.66.
total df in the k samples is N - 1 = 19
treatment df is k – 1 = 3
error df is k(n - 1) = N - k = 16
Using the F tables in our text book,
p < .01.
• One-tailed test of nondirectional
hypothesis
Deviation Method
• SSTOT =  (Yij - GM)2
= (1 - 5)2 + (2 - 5)2 +...+ (9 - 5)2 = 138.
• SSA =  [nj  (Mj - GM)2]
• SSA = n   (Mj - GM)2 with equal n’s
= 5[(2 - 5)2 + (3 - 5)2 + (7 - 5)2 + (8 - 5)2] = 130.
• SSE =  (Yij - Mj)2
= (1 - 2)2 + (2 - 2)2 + .... + (9 - 8)2 = 8.
Computational Method
SSTOT
2
G
 Y 2 
N
= (1 + 4 + 4 +.....+ 81) - [(1 + 2 + 2 +.....+ 9)2]
 N = 638 - (100)2  20 = 138.
Tj2
2
G
SS A  

nj
N
SS A 
 T j2
n
G2

N
= [102 + 152 + 352 + 402]  5 - (100)2  20 =
130.
SSE = SSTOT – SSA = 138 - 130 = 8.
Source Table
Source
SS
df
Teaching Method
130
3
Error
8
16
Total
138
19
MS
F
43.33 86.66
0.50
p
< .001
Magnitude of Effect
SSA 130
 

 .942
SSTot 138
2
• Omega Square is less biased
SSA  (k  1)  MSE
 
 .927
SSTOT  MSE
2
Magnitude of Effect
• Put a confidence interval on eta-squared.
• http://core.ecu.edu/psyc/wuenschk/SPSS/
SPSS-Programs.htm
Magnitude of Effect
• Enter F and df into NoncF.sav
Magnitude of Effect
• Run syntax file NoncF3.sps.
• CI runs from .837 to .960.
Pairwise Comparisons and Familywise Error
• fw is the alpha familywise, the
conditional probability of making one or
more Type I errors in a family of c
comparisons.
• pc is the alpha per comparison, the
criterion used on each individual
comparison.
• Bonferroni: fw  cpc
c = 6, pc = .01
• alpha familywise might be as high as
6(.01) = .06.
• What can we do to lower familywise error?
Fisher’s Procedure
• Also called the “Protected Test” or
“Fisher’s LSD.”
• Do ANOVA first.
• If ANOVA not significant, stop.
• If ANOVA is significant, make pairwise
comparisons with t.
• For k = 3, this will hold familywise error at
the nominal level, but not with k > 3.
Computing t
• Assuming homogeneity of variance, use
the pooled error term from the ANOVA:
t
Mi  M j
1
1 

MSE 

n

n
i
j


• For A versus D:
t (16)  (8  2)  .2  13.416, p  .001
• For A versus C and B versus D:
t (16)  5  .2  11.180, p  .001
• For B versus C
t (16)  (7  3)  .2  8.944, p  .001
• For A vs B, and C vs D,
t (16)  1 .5(1/ 5  1/ 5)  2.236, p  .04
Underlining Means Display
• arrange the means in ascending order
• any two means underlined by the same
line are not significantly different from one
another
Group A
B
C
D
Mean 2
3
7
8
The Bonferroni Procedure
• Does NOT require that ANOVA be
conducted or, if conducted, that ANOVA be
significant.
• Compute an adjusted criterion of
significance to keep familywise error at
desired level
 pc 
 fw
c
• For our data,
.01
 
 pc
 .00167
6
• Compare each p with the adjusted criterion.
• For these data, we get same results as with
Fisher’s procedure.
• In general, this procedure is very conservative
(robs us of power).
Ryan-Einot-Gabriel-Welsch Test
• Does not require a significant ANOVA.
• Holds familywise error at the stated level.
• Has more power than other techniques
which also adequately control familywise
error.
• SPSS will do it for you.
Which Test Should I Use?
• If k = 3, use Fisher’s Procedure
• If k > 3, use REGWQ
• Remember, ANOVA does not have to be
significant to use REGWQ.
APA-Style Presentation
Table 1
Effectiveness of Four Methods of Teaching ANOVA
Method
M
SD
Ancient
2.00A
.707
A
Backwards
3.00
.707
B
Computer-Based
7.00
.707
Devoted
8.00B
.707
Note. Means with the same letter in their
superscripts do not differ significantly from one
another according to a Bonferroni test with a .01
limit on familywise error rate.
Teaching method significantly affected
test scores, F(3, 16) = 86.66, MSE = 0.50,
p < .001, 2 = .942, CI.95 = .837, .960. Pairwise
comparisons were made with Bonferroni
tests, holding familywise error rate at a
maximum of .01. As shown in Table 1, the
computer-based and devoted methods
produced significantly better student
performance than did the ancient and
backwards methods.
Computing ANOVA From Group Means and
Variances with Unequal Sample Sizes
Semester
Mean
SD
N
p
Spring 89
4.85
.360
34
34/133 = .2556
Fall 88
4.61
.715
31
31/133 = .2331
Fall 87
4.61
.688
36
36/133 = .2707
Spring 87
4.38
.793
32
32/133 = .2406
j
pj 
nj
N
MSE   p j s 2j  .2556(.360)2  .2331(.715)2  .2707(.688)2  .2406(.793)2  .4317.
GM =  pj Mj =.2556(4.85) + .2331(4.61) + .2707(4.61) + .2406(4.38) = 4.616.
Among Groups SS = 34(4.85 - 4.616)2 + 31(4.61 - 4.616)2 + 36(4.61 - 4.616)2
+ 32(4.38 - 4.616)2 = 3.646.
With 3 df, MSA = 1.215, and F(3, 129) = 2.814, p = .042.