Transcript L643: Evaluation of Information Systems

S519: Evaluation of
Information Systems
Social Statistics
Inferential Statistics
Chapter 11: ANOVA
Last week
This week
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When to use F statstic
How to compute and interpret
Using FTEST and FDIST functions
How to use the ANOVA
Analysis of variance
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Goudas, M.; Theodorakis, Y.; and Karamousalidis, G.
(1998). Psychological skills in basketball: Preliminary
study for development of a Greek form of the Athletic
Coping Skills Inventory. Perceptual and Motor Skills, 86,
59-65
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Group 1: athletes with 6 years of experience or less
Group 2: athletes with 7 to 10 years of experience
Group 3: athletes with more than 10 years of experience
The athletes are not being tested more than once.
One factor: psychological skills (experiences)
Which statistic test should we use?
Simple analysis of variance
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There is one factor or one treatment variable
being explored and there are more than two
levels within this factor.
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Simple ANOVA: one-way analysis of variance
or single factor
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It tests the difference between the means of more
than two groups on one factor or dimension.
Simple ANOVA
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Any analysis where
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There is only one dimension or treatment or one
variable
There are more than two levels of the grouping
factor, and
One is looking at differences across groups in
average scores
Using simple ANOVA (F test)
F value
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Logic: if there are absolutely no variability within each
group (all the scores were the same), then any
difference between groups would be meaningful.
ANOVA: compares the amount of variability between
groups (which is due to the grouping factor) to the
amount of variability within groups (which is due to
chance)
MSbetween between groupvariable
F
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MSwithin
within  groupvariable
F value
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F=1
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F increase
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The amount of variability due to within-group differences is equal
to the amount of variability due to between-group differences 
any difference between groups would not be significant
The average different between-group gets larger the difference
between groups is more likely due to something else (the
grouping factor) than chance (the within-group variation)
F decrease
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The average different between-group gets smaller the
difference between groups is more likely due to chance (the
within-group variation) rather than due to other reasons (the
grouping factor)
Example
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Three groups of preschoolers and their
language scores, whether they are overall
different?
Group 1 Scores
Group 2 Scores
87
86
76
56
78
98
77
66
75
67
Group 3 Scores
87
89
85
91
99
96
85
87
79
89
81
90
82
89
78
96
85
96
91
93
F test steps
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Step1: a statement of the null and research
hypothesis
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One-tailed or two-tailed (there is no such thing in
ANOVA)
H 0 : 1  2  3
H1 : X 1  X 2  X 3
F test steps
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Step2: Setting the level of risk (or the level of
significance or Type I error) associated with
the null hypothesis
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0.05
F test steps
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Step3: Selection of the appropriate test
statistics
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See Figure 11.1 (S-p227)
Simple ANOVA
F test steps
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Step4: Computation of the test statistic value
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the between-group sum of squares = the sum of
the differences between the mean of all scores
and the mean of each group score, then squared
The within-group sum of squares = the sum of the
differences between each individual score in a
group and the mean of each group, then squared
The total sum of square = the sum of the
between-group and within-group sum of squares
F test steps
Group 1 Scores
87
86
76
56
78
98
77
66
75
67
x square
Group 2 Scores
7569
87
7396
85
5776
99
3136
85
6084
79
9604
81
5929
82
4356
78
5625
85
4489
91
n
∑x
10
766
10
852
10
916
X
76.6
85.2
91.6
59964
58675.6
72936
72590.4
84010
83905.6
( X 2 )
( X ) 2 / n
x square
7569
7225
9801
7225
6241
6561
6724
6084
7225
8281
Group 3 Scores
89
91
96
87
89
90
89
96
96
93
x square
7921
8281
9216
7569
7921
8100
7921
9216
9216
8649
N
∑∑X
( X ) 2 / N
 (X ) 2
 ( X ) 2 / n
30
2534
214038.5333
216910
215171.6
F-test
Between sum of
squares
within sum of
squares
total sum of
squares
 ( X ) / n  ( X )
 ( X )   ( X ) / n
2
2
 ( X )
2
/N
215171.6-214038.53
1133.07
216910-215171.60
1738.40
216910-214038.53
2871.47
2
2
 ( X ) 2 / N
F test steps
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Between-group degree of freedom=k-1
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k: number of groups
Within-group degree of freedom=N-k
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N: total sample size
source
Between
groups
sums of
squares
mean sums of
squares
df
1133.07
2
566.53
Within gruops
1738.40
27
64.39
Total
2871.47
29
F
8.799
F test steps
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Step5: determination of the value needed for
rejection of the null hypothesis using the
appropriate table of critical values for the
particular statistic
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Table B3 (S-p363)
df for the denominator = n-k=30-3=27
df for the numerator = k-1=3-1=2
3.36
F test steps
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Step6: comparison of the obtained value and
the critical value
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If obtained value > the critical value, reject the null
hypothesis
If obtained value < the critical value, accept the
null hypothesis
8.80 and 3.36
F test steps
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Step7 and 8: decision time
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What is your conclusion? Why?
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How do you interpret F(2, 27)=8.80, p<0.05
Excel: ANOVA
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Three different ANOVA:
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Anova: single factor
Anova: two factors with replication
Anova: two factors without replication
ANOVA: a single factor
Anova: Single Factor
SUMMARY
Groups
Group 1 Scores
Group 2 Scores
Group 3 Scores
Count
10
10
10
ANOVA
Source of Variation
SS
Between Groups
1133.067
Within Groups
1738.4
Total
2871.467
Sum
Average Variance
766
76.6 143.1556
852
85.2
38.4
916
91.6
11.6
df
MS
F
P-value
F crit
2 566.5333 8.799126 0.001142 3.354131
27 64.38519
29
Exercise
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S-p241
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