In: Statistics and Probability
Complete the following table and test whether there is a difference among the group means at a 0.025 level of significance.
Source | df | SS | MS | F |
---|---|---|---|---|
Among Groups | 787.5 | 3.5 | ||
Within Groups | 16 | 1200 | ||
Total |
Answer:
Given ANOVA table is
Source | df | SS | MS | F |
Among Groups | 787.5 | 3.5 | ||
Within Groups | 16 | 1200 | ||
Total |
ANOVA table is :
Source | df | SS | MS | F |
Among Groups | 3 | 787.5 | 262.5 | 3.5 |
Within Groups | 16 | 1200 | 75 | |
Total | 19 | 1987.5 |
Here, H0: there is no difference among the group means.
H1: there is a difference among the group means
Now the critical value is
F(crit) = 4.08 ...( from f distribution table with df n1=3, n2=16 and alpha=0.025)
Also from Excel
=FINV(0.025,3,16) = 4.0768 = 4.08 ( rounded to 2 decimals)
Conclusion:
We know if F> F(crit) then we reject null hypothesis.
As F = 3.5 is less than F(crit) = 4.08, we can not reject the null hypothesis.
Hence we conclude that there is a difference among the group means at a 0.025 level of significance.