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 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

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 (x)2 and x2.

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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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