Biostatistics. Unit 8 ANOVA. 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 . ANOVA—Analysis of Variance.
Biostatistics Unit 8 ANOVA
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.
ANOVA—Analysis of Variance ANOVA was developed in the 1920s by Ronald A. Fisher (1890-1962) who worked for the British Government Agricultural Department.
Data Table • Data for ANOVA are placed in a data table. • There must be at least three groups of data.
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
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.
Calculations • 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.
Calculations 2. An ANOVA table is prepared which includes: df Degrees of freedom SS Sum of squares MS Mean squares F Variance ratio
Calculations 3. N and k are used to calculate degrees of freedom. TOTAL df = N – 1 GROUP df = k – 1 ERROR df = N - k
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.)
Sample ANOVA Calculations a. Given Opercular breathing rates of goldfish at different temperatures.N = 48 (number of measurements) k = 6 (number of groups)
Sample ANOVA Calculations b. Assumptions • normal distribution of data • independent simple random samples • constant variance
Sample ANOVA Calculations c. Hypotheses H0: all means are equal HA: not all the means are equal
Sample ANOVA Calculations d. Statistical test
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.
Sample ANOVA Calculations [A] Calculate correction factor Remember: there is a big difference between (Sx)2 and Sx2.
Sample ANOVA Calculations [B] Calculate SS Total
Sample ANOVA Calculations [C] Calculate SS Group
Sample ANOVA Calculations [D] Calculate SS Error
Sample ANOVA Calculations [E] Calculate MS Group
Sample ANOVA Calculations [F] Calculate MS Error
Sample ANOVA Calculations [G] Calculate Variance Ratio Result: the completed ANOVA table
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.
Sample ANOVA Calculations g. Conclusions We conclude that not all the means of the groups are equal.