biostatistics unit 8 anova 1. anova—analysis of variance anova is used to determine if there is...
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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 x, x2 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. An ANOVA table is prepared which includes:
df Degrees of freedom
SS Sum of squares
MS Mean squares
F 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
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 (x)2 and x2.
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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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