In: Statistics and Probability
MA1 = 102.5
MA2 = 103
MA3 = 104.5
Source |
SS |
df |
MS |
F |
||
Factor A |
17.333 |
2 |
8.667 |
7.00 |
||
Persons |
16.667 |
7 |
2.380953 |
|||
Error (within) |
17.333 |
14 |
1.238 |
|||
Total |
51.33 |
23 |
Null Hypothesis H0: The means of caffeine consumptions are equal for all groups.
Alternative Hypothesis Ha: Not all of the means of caffeine consumptions are equal.
F Test statistic for Factor A = 7.00
Critical value of F at α = .05 and df = 2, 14 is 3.74
Since the observed test statistic (7.00) is greater than the critical value (3.64) , we reject H0 and conclude that there is statistically significant difference between the means of caffeine consumption.
η2 = SS Factor A / SS Total = 17.333 / 51.33 = 0.3377
Using Tukey's PostHoc test,
Tukey HSD is calculated using the below formula
where is a critical value of the studentized range for α, the number of treatments or samples r, and the within-groups degrees of freedom . We get this value from studentized range table.
is the within groups mean square from the ANOVA table and n is the sample size for each treatment.
N = df Total + 1 = 23 + 1 = 24
n = N / r = 24 / 3 = 8
From Anova table,
= 2 , = 14
n = 8, r = 3, α = 0.05
From studentized range table, = 3.701
So,
So, any aboslute mean difference greater than 1.8505 is significantly different.
A2 - A1 = 103 - 102.5 = 0.5
A3 - A1 = 104.5 - 102.5 = 2
A3 - A2 = 104.5 - 103 = 1.5
Only the difference between A1 and A3 is greater than the critical Tukey HSD value. Thus, the means of A1 and A3 are significanlty different at α = .05