Biostatistics
Unit 8
ANOVA
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ANOVA—Analysis of Variance
• ANOVA is used to determine if there is any
significant difference between the means of
groups of data.
• In one-way ANOVA these groups vary under
the influence of a single factor.
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ANOVA—Analysis of Variance
ANOVA was developed in the 1920s by
Ronald A. Fisher (1890-1962) who worked for
the British Government Agricultural
Department.
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Data Table
• Data for ANOVA are placed in a data
table.
• There must be at least three groups of
data.
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Assumptions and Hypotheses
The assumptions in ANOVA are:
-normal distribution of the data
-independent simple random samples
-constant variance
The hypotheses are:
H0: all means are equal
HA: not all the means are equal
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Test Statistic
• The test statistic is V.R. which is
distributed as F with the appropriate
number of numerator degrees of
freedom and denominator degrees of
freedom.
• A large value of F indicates rejection of
H0.
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Calculations
1. Basic calculations are done to determine
the values of Sx, Sx2 and n for each group.
Place the data from each group in one of the
lists of the TI-83. Use of 1-Var Stats gives
these values automatically.
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Calculations
2.
df
SS
MS
F
An ANOVA table is prepared which includes:
Degrees of freedom
Sum of squares
Mean squares
Variance ratio
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Calculations
3. N and k are used to calculate
degrees of freedom.
TOTAL df = N – 1
GROUP df = k – 1
ERROR df = N - k
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Calculations
4.
Calculations for ANOVA table values:
[A] correction factor
[D] SS Error
[B] Sum of Squares Total [E] MS Group
Value (SS Total)
[F] MS Error
[C] SS Group
[G] F (V.R.)
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Sample ANOVA Calculations
a. Given
Opercular breathing rates of goldfish
at different temperatures.
N = 48 (number of measurements)
k = 6 (number of groups)
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Sample ANOVA Calculations
b. Assumptions
• normal distribution of data
• independent simple random samples
• constant variance
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Sample ANOVA Calculations
c. Hypotheses
H0: all means are equal
HA: not all the means are equal
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Sample ANOVA Calculations
d.
Statistical test
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Sample ANOVA Calculations
Decision criteria
The critical value of F with 5 numerator
degrees of freedom and 42 denominator
degrees of freedom is about 2.45 at the
95% confidence level. We reject H0 if
V.R. > 2.45.
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Sample ANOVA Calculations
[A] Calculate correction factor
Remember: there is a big difference
between (Sx)2 and Sx2.
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Sample ANOVA Calculations
[B] Calculate SS Total
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Sample ANOVA Calculations
[C] Calculate SS Group
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Sample ANOVA Calculations
[D] Calculate SS Error
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Sample ANOVA Calculations
[E] Calculate MS Group
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Sample ANOVA Calculations
[F] Calculate MS Error
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Sample ANOVA Calculations
[G] Calculate Variance Ratio
Result: the completed ANOVA table
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Sample ANOVA Calculations
f.
Discussion
• The 95% CI level for F with 5 numerator
degrees of freedom and 42 denominator
degrees of freedom is 2.45 as read from the
F tables.
• The actual value is 12.01 with a probability
of 2.98 x 10-7.
• This means that H0 is rejected.
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Sample ANOVA Calculations
g. Conclusions
We conclude that not all the means of the
groups are equal.
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fin
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