For this week’s forum, locate a research article that used an ANOVA and apost-hoc analysis in their methods/results.. Please attach the article or provide a link so classmates can view it. Describe how the ANOVA and post-hoc analysis was used to answer the research (i.e. what did it compare) and where the group differences were found in the post-hoc. Please share the levels of the dependent and independent variable(s). As always, please select an article that no one else has used or discussed for this week.
SPHS500 – Statistics for
Sports and Health Sciences
Week 5
Objectives
• Analysis of Variance (ANOVA)
• One-way ANOVA
• Repeated Measures ANOVA
• Interpreting test results
• Post-hoc tests
ANOVA
• ANOVA is used when more than two groups are compared
• In order to conduct an ANOVA, several assumptions must be made:
• The population from which the samples are drawn are normally distributed
• The populations from which the samples are drawn have equal variances
• Some ask – why not multiple t-tests?
• That increases the likelihood of a Type I error
Decision Table for Inferential Statistic
Normal deviate z-test
One sample t-test
Independent t-test
Dependent t-test
One-way ANOVA and
post hoc test
Not covered
Factorial ANOVA
Not covered
Chi-square goodness of fit
Chi-square independence
Pearson r
Not covered
One Way (Between subjects) ANOVA
• Single independent variable
• IV = depression level
• 3 levels (medication, counseling, diet supplement)
ANOVA
Analysis of Variance:
• Used with a quantitative Dependent Variable
• Comparison among 3 or more groups
• A Between subjects (one-way) ANOVA is when there are 3 different groups
• Freshman, sophomore, junior, senior
• Categories of an Independent Variable (Low, Medium, High)
• A Repeated measures ANOVA is when the same group is compared at
different time points (measures are repeated)
• Baseline, 3-month, 6-month, 9-month, post
Types of variance in an ANOVA:
1. Between-groups variance
• Variance due to differences between the groups (i.e., between the
factors/levels).
2. Within-groups variance
• Variance due to differences within the groups (i.e., between the individuals).
3. Total variability
• The difference of each score from the grand mean (actually the squared
difference)
The F-Ratio
Between-subjects variability
F=
Within-subjects variability
Treatment effect + Indiv. Diff. + Exper. Error
F=
Indiv. Diff. + Exper. Error
The F-ratio
F=
Between-subjects variability
Within-subjects variability
Variance = Mean Square (MS) in ANOVA since variance is the
mean of the squared deviation scores
• For each source find:
• Sum of squares (Between-groups, Within-groups, Total)
• Degrees of freedom
• Compute Mean Sum of Squares
• F = MSB / MSW = (SSB / dfB) / (SSW / dfW)
Let’s walk through an example:
Do mini-golfers play better (i.e. lower score) using
a white, blue or red ball (3 different groups)?
Red
White
Blue
3
6
4
5
3
8
2
7
7
3
1
5
4
4
5
• What are your null and alternate hypotheses?
Our hypotheses
• Null hypothesis – there is no difference between the
means for the three different groups
• Ho: μ1 = μ2 = μ3
• example, Group 1 = Group 2 = Group 3
• Alternate hypothesis
• ANOVA is omnibus – tests for an overall difference between
means (neither one nor two-tailed)
• HA: X1 ≠ X2 ≠ … ≠ Xk
• HA: at least one of the means is different
Compute Sum of Squares
Red Ball (n1=5)
White Ball (n2=5)
Blue Ball (n3=10)
TOTAL (N=15) (k=3)
3
6
4
5
3
8
2
7
7
3
1
5
4
4
5
X =3.4
X = 4.2
X = 5.8
∑X = 17
∑X = 21
∑X = 29
∑(∑X) = 67
∑(X2) = 63
∑(X2) = 111
∑(X2) = 179
∑(∑ (X2)) = 353
Compute Sum of Squares
Red Ball (n1=5)
White Ball (n2=5)
Blue Ball (n3=5)
TOTAL (N=15) (k=3)
X = 3.4
X = 4.2
X = 5.8
∑X = 17
∑X = 21
∑X = 29
∑(∑X) = 67
∑(X2) = 63
∑(X2) = 111
∑(X2) = 179
∑(∑ (X2)) = 353
67 ) 2
((X )) 2
(2534
353
53.7 .5
SStot = (( X )) −
= 216910
−
= 2871
N
1530
sum of the scores in the column)2
((X )) 2
SSb = (
)−(
)
n of the scores in the column
N
21 2
29 2
53.7 ) 2
17 2
(766)
(852)
(916)
(2534
= (
+
+
)−(
)
10
30
510
15
5
510
= 1133.1
14.93
2
53.7 – 1133.1
14.93
38.8
SSw = SStot – SSb = 2871.5
= 1738.4
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