### Chapter 1: Introduction to Statistics

```COURSE: JUST 3900
TIPS FOR APLIA
Chapter : 12
Analysis of Variance: ANOVA
Developed By:
John Lohman
Michael Mattocks
Aubrey Urwick
Key Terms: Don’t Forget
Notecards
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Factors (p. 388)
Levels (p. 388)
Testwise Alpha Level (p. 391)
Experimentwise Alpha Level (p. 391)
Error Term (p. 394)
Post Hoc Tests or Post Tests (p. 416)
ANOVA Notation
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k is used to identify the number of treatment conditions
n is used to identify the number of scores in each treatment
condition
N is used to identify the total number scores in the entire study
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T stands for treatment total and is calculated by ∑X, which equals
the sum of the scores for each treatment condition
G stands for the sum of all scores in a study (Grand Total)
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N = kn, when samples are the same size
Calculate by adding up all N scores or by adding treatment total (G=∑T)
You will also need SS and M for each sample, and ∑X2 for the entire
set of all scores.
Formulas
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F-ratio:  =

ℎ
2
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SStotal:  =
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SSwithin: ℎ = 1 + 2 + 3 …
SSbetween:  =  − ℎ
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−
2

2

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SSbetween:  =
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dftotal:  =  − 1
dfwithin: ℎ =  − 1 or ℎ =  −
dfbetween:  =  − 1
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−
2

More Formulas
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2
MSwithin: ℎ = ℎ
=
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MSbetween:  =
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Tukey’s HSD:  =
ℎ
ℎ
2

=

ℎ


Scheffe Test:    =

Effect Size: η2 =

ℎ

+ℎ
=

Hypothesis Testing with ANOVA
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Question 1: A psychologist studied three computer
keyboard designs. Three samples of individuals were
given material to type on a particular keyboard, and the
number of errors committed by each participant was
recorded. The data are as follows:
Keyboard A
Keyboard B
Keyboard C
0
6
6
4
8
5
0
5
9
1
4
4
0
2
6
T=
SS =
T=
SS =
T=
SS =
N=
G=
ΣX2 =
Hypothesis Testing with ANOVA
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Question 1: Are these data sufficient to conclude that
there are significant differences in typing performance
among the three keyboard designs? Set alpha at
α = 0.05
Keyboard A
Keyboard B
Keyboard C
0
6
6
4
8
5
0
5
9
1
4
4
0
2
6
T=5
SS = 12
T = 25
SS = 20
T = 30
SS = 14
N = 15
G = 60
ΣX2 = 356
Hypothesis Testing with ANOVA
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Step 1: State the hypothesis.
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H0: μ1 = μ2 = μ3 (Type of keyboard has no effect)
H1: At least one of the treatment means is different.
Hypothesis Testing with ANOVA
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Step 2: Locate the critical region
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=  − 1 = 3 − 1 = 2
ℎ =  −  = 15 − 3 = 12
For this problem df = 2,12 and the critical value for α = 0.05 is
F = 3.88.
If F-ratio ≤ Fcritical (3.88), then fail to reject H0.
If F-ratio > Fcritical (3.88), then reject H0.
Hypothesis Testing with ANOVA
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Step 3: Perform the analysis.
2 −
2

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ℎ = 1 + 2 + 3 = 12 + 20 + 14 = 46
=  − ℎ = 116 − 46 = 70
or  =
2

2
−
=
52
5
+
252
5
+
302
5
−
602
15
= 356 −
3600
15
=
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= 356 −
602
15
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= 310 −

70
2
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2
=
=  =
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2
ℎ = ℎ
= ℎ = 12 = 3.83
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=

46
ℎ

ℎ
35
= 3.83 = 9.14
602
15
= 35
= 356 − 240 = 116
= 310 − 240 = 70
Hypothesis Testing with ANOVA
Sources
SS
df
MS
Between
70
2
35
Within
46
12
3.83
Total
116
14
F = 9.14
Hypothesis Testing with ANOVA
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Step 4: Make a decision
If F-ratio ≤ Fcritical (3.88), then fail to reject H0.
If F-ratio > Fcritical (3.88), then reject H0.
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F-ratio (9.14) > Fcritical (3.88). Therefore, we reject H0. The type of
keyboard used has a significant effect on the number of errors
committed.
Computing Effect Size for
ANOVA
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Question 2: Compute effect size (η2), the percentage of
variance explained, for the data that were analyzed in
Question 1.
Computing Effect Size for
ANOVA
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η2 =

+ℎ
=

=
70
116
= 0.60 = 60%
Post Hoc Tests
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Question 3: For the data used in Question 1, perform a
post hoc test to determine which mean differences are
significant and which are not. Use both Tukey’s HSD and
the Scheffe Test.
Post Hoc Tests: Tukey’s HSD
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=
ℎ

1.
Find q. q = 3.77 (Table B.5, p.708)
2.
=
3.
Thus, the mean difference between any two samples must be at
least 3.23 to be significant.
Find the means for each treatment.
4.
1.
=
2.
=
3.
=
ℎ

= 3.77
5
=5=1
=
=
25
5
30
5
=5
=6
3.83
5
= 3.77 0.766 = 3.23
Post Hoc Tests: Tukey’s HSD
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=
6.
7.
8.
ℎ

HSD = 3.23
−  = 1 − 5 = −4, Treatment A is significantly different than
Treatment B.
−  = 1 − 6 = −5, Treatment A is significantly different than
Treatment C.
−  = 5 − 6 = −1, Treatment B is not significantly different
than Treatment C.
Post Hoc Tests: Scheffe Test
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First, compute SSbetween for Treatments A and B.
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=
2

2
−
=
52
5
+
252
5
−
302
10
= 5 + 125 − 90 = 40
Notice: G is equal to the total of Treatments A and B, not A, B, and C.
Similarly, N is equal to nA + nB.
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Now, find MSbetween.
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=

=
40
2
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=

ℎ
= 3.83 = 5.22
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For df (2,12) and α = 0.05, the critical region for F is 3.88. Therefore
our obtained F-ratio is in the critical region, and we must conclude
that these data show a significant difference between treatment A
and treatment B.
= 20
For dfbetween, use k-1.
20
Post Hoc Tests: Scheffe Test
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First, compute SSbetween for Treatments A and C.
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=
2

2
−
=
52
5
302
5
+
−
352
10
= 5 + 180 − 122.5 = 62.5
Notice: G is equal to the total of Treatments A and C, not A, B, and C.
Similarly, N is equal to nA + nC.
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Now, find MSbetween.
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=

=
62.5
2
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=

ℎ
=
31.25
3.83
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For df (2,12) and α = 0.05, the critical region for F is 3.88. Therefore
our obtained F-ratio is in the critical region, and we must conclude
that these data show a significant difference between treatment A
and treatment C.
= 31.25 For dfbetween, use k-1.
= 8.16
Post Hoc Tests: Scheffe Test
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First, compute SSbetween for Treatments B and C.
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=
2

2
−
=
252
5
+
302
5
552
− 10
= 125 + 180 − 302.5 = 2.5
Notice: G is equal to the total of Treatments B and C, not A, B, and C.
Similarly, N is equal to nB + nC.
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Now, find MSbetween.
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=

=
2.5
2
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=

ℎ
= 3.83 = 0.33
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For df (2,12) and α = 0.05, the critical region for F is 3.88. Therefore
our obtained F-ratio is not in the critical region, and we must
conclude that these data show no significant difference between
treatment B and treatment C.
= 1.25
For dfbetween, use k-1.
1.25
Assumptions for ANOVA
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Question 4: What three assumptions are required for
ANOVA?
Assumptions for ANOVA
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