Ryan S. answered 09/30/13
Tutor
4.8
(10)
Mathematics and Statistics
Here is the output from the Excel Add-in with the alpha set to .05
SUMMARY
Groups Count Sum Average Variance
Group 1 6 625 104.1666667 44.16666667
Group 2 23 2507 109 344.9090909
Group 3 11 1300 118.1818182 267.5636364
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 937.930303 2 468.9651515 1.654991727 0.204926644 3.251923846
Within Groups 10484.4697 37 283.3640459
Total 11422.4 39
Notice the F value, P-value, and F critical
F is the observed value of the F statistic.
F critical is the value of the F statistic from the F distribution with numerator df=2 and denominator df=39 where Pr(F>=F critical) = .05 = alpha.
P-value is Pr(F>= F observed ).
So the probability of observing our F statistic or greater (20.5%) is much higher than than our significance level (5%). Equivalently our F observed (1.65) is much less than F critical (3.25).
Essentially the data shows that there is too high a probability that the difference between the groups is due to variability rather than treatment differences to reject the null hypothesis.
The conclusion is not to reject H0 that the group means are equal.
SUMMARY
Groups Count Sum Average Variance
Group 1 6 625 104.1666667 44.16666667
Group 2 23 2507 109 344.9090909
Group 3 11 1300 118.1818182 267.5636364
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 937.930303 2 468.9651515 1.654991727 0.204926644 3.251923846
Within Groups 10484.4697 37 283.3640459
Total 11422.4 39
Notice the F value, P-value, and F critical
F is the observed value of the F statistic.
F critical is the value of the F statistic from the F distribution with numerator df=2 and denominator df=39 where Pr(F>=F critical) = .05 = alpha.
P-value is Pr(F>= F observed ).
So the probability of observing our F statistic or greater (20.5%) is much higher than than our significance level (5%). Equivalently our F observed (1.65) is much less than F critical (3.25).
Essentially the data shows that there is too high a probability that the difference between the groups is due to variability rather than treatment differences to reject the null hypothesis.
The conclusion is not to reject H0 that the group means are equal.