ANOVA Statistics Evaluation - Assignment Example

As stated in the question the one-way ANOVA with 4 groups is statistically significant. This suggests that the means are all not equal and at least one group is different from another. However, the ANOVA test does not say anything about which groups are different from each other. We cannot use t-test skills here because then we have to do 6 separate t-tests to know which groups are different from each other. This will also result in increasing the probability of making a Type I error. Suppose, the selected level of significance, α is .05. The probability of a Type I error on any single comparison is .05. However, the probability of a Type I error on multiple comparisons will be higher. For 6 comparisons, Type I error will be: Type I error = 1 – (1 – .05)c = 1 – (1 – .05)6 = 1 – .735 = .265.

This means 26.5% of the time we will reject the null hypothesis of equal means in favor of the alternate hypothesis even when the null hypothesis is true.

Therefore, we will need to perform a post hoc test to determine which groups are different from each other. One such post hoc test is Tukey’s HSD post hoc test, which allows one to make pairwise comparisons among the sample means while maintaining an acceptable alpha (usually.05) when the groups have equal sample sizes, n. If groups have unequal sample sizes, then another post hoc test, such as Fisher’s protected t-test is used for multiple comparisons (Jackson, 2012). In Tukey’s HSD post hoc test, absolute mean differences are compared with the critical range to determine the difference between any pair of means. The formula for Tukey’s HSD is: HSD.05 = Q(k, dfwithin).

The value of the Q is taken from the  Tukey Q Test table. If the absolute mean difference of any pair of groups is equal to or greater than the calculated HSD value, then there is a significant difference between the groups at the .05 level.