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Biostatistics Unit 8 ANOVA 1

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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. 2

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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. 3

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Data Table Data for ANOVA are placed in a data table. There must be at least three groups of data. 4

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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: H 0 : all means are equal H A : not all the means are equal 5

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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 H 0. 6

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Calculations 1.Basic calculations are done to determine the values of x, x 2 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. 7

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Calculations 2.An ANOVA table is prepared which includes: dfDegrees of freedom SSSum of squares MSMean squares FVariance ratio 8

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Calculations 3.N and k are used to calculate degrees of freedom. TOTALdf = N – 1 GROUP df = k – 1 ERROR df = N - k 9

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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.) 10

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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) 11

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Sample ANOVA Calculations b.Assumptions normal distribution of data independent simple random samples constant variance 13

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Sample ANOVA Calculations c.Hypotheses H 0 : all means are equal H A : not all the means are equal 14

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Sample ANOVA Calculations d.Statistical test 15

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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 H 0 if V.R. > 2.45. 16

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Sample ANOVA Calculations [A]Calculate correction factor Remember: there is a big difference between ( x) 2 and x 2. 18

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Sample ANOVA Calculations [B]Calculate SS Total 19

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Sample ANOVA Calculations [C]Calculate SS Group 20

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Sample ANOVA Calculations [D]Calculate SS Error 21

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Sample ANOVA Calculations [E]Calculate MS Group 22

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Sample ANOVA Calculations [F]Calculate MS Error 23

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Sample ANOVA Calculations [G]Calculate Variance Ratio Result: the completed ANOVA table 24

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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 H 0 is rejected. 25

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Sample ANOVA Calculations g.Conclusions We conclude that not all the means of the groups are equal. 26

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