fessor Johnston conducted an experiment where subjects either drank 5, 3, or 0 c

Question

fessor Johnston conducted an experiment where subjects either drank 5, 3, or 0 cups of coffee and then measured their alertness. The first two conditions had 55 subjects but the cont condition had 51 subjects. Professor Johnston obtained a significant ANOVA. Is he now justified in using a Tukey's HSD test? rol a. This can't be answered because we don't know if the one-way ANOVA was significant. b. Yes, because there are three levels within the one-way ANOVA, we must conduct the Tukey HSD test. c. No, because the results of the ANOVA are not significant. d. No, because the subjects in all the conditions are not all equal.

Explanation / Answer

ans ->

No, because the subjects in all the conditions are not all equal.

The Tukey Test (or Tukey procedure), also called Tukey’s Honest Significant Difference test, is a post-hoc test based on the studentized range distribution. An ANOVA test can tell you if your results are significant overall, but it won’t tell you exactly where those differences lie. After you have run an ANOVA and found significant results, then you can run Tukey’s HSD to find out which specific groups’s means (compared with each other) are different. The test compares all possible pairs of means.

But Tukey HSD uses with Tukey-Kramer formula when treatments (sample groups) have unequal observations (i.e. unbalanced observations)

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