One Way Between Subjects Analysis of Variance

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Inference of a Relationship Using One-Way ANOVA (continued)

QUESTION 2: WHAT’S THE STRENGTH OF THE RELATIONSHIP?

- We wanted to know how much of an influence different amounts of caffeine had on anagram performance.

In other words, how much of the total variability in anagram performance (SS_total) is due to the fact that people drank different amounts of caffeine (SS_between)?

- We measure this with eta squared:

QUESTION 3: WHAT’S THE NATURE OF THE RELATIONSHIP?

- Now we know that there is a statistically significant effect of caffeine on anagram performance, and it’s a large effect – what’s the nature of the effect?

- Does more caffeine lead to improved performance? Or decreased performance? Or does it improve performance up to a point, then decrease performance?

- To answer this question, we have to compare the group means to each other.

- We need to make a comparison of each group mean to each of the other group means:

Comparison

- When we reject Null (statistically significant results), we are saying there are some differences, but we don’t know which means are statistically significantly different.

Could do 3 t-tests, but you have Type I = .05 for each test (Familywise alpha = .15)

So people use another statistical test that uses a lower alpha, called TUKEY’S HSD to make "all possible pairwise comparisons."

TUKEY’S HSD:

- The first step is to compute the absolute difference between each pair of group means.

Comparison

Absolute Difference

|69.00 – 82.00| = 13.00

|69.00 – 62.00| = 7.00

|82.00 – 62.00| = 20.00

- We determine if this difference is statistically significant by comparing it to a "critical difference" (CD), using:

- We already know MSwithin and n – we get q (Studentized range value) from Appendix G (p. 603):

look up q value using df denominator and k (# levels of IV)

- Use these values to calculate CD.

Comparison	Absolute Difference	Critical Difference
	\|69.00 – 82.00\| = 13.00	5.105
	\|69.00 – 62.00\| = 7.00	5.105
	\|82.00 – 62.00\| = 20.00	5.105

- Now we just compare our absolute difference to the critical difference – if higher, reject null, if lower, do not reject null for that comparison

Comparison	Absolute Difference	Critical Difference	Statistically Significant?
	\|69.00 – 82.00\| = 13.00	5.105	Yes
	\|69.00 – 62.00\| = 7.00	5.105	Yes
	\|82.00 – 62.00\| = 20.00	5.105	Yes

- So our Tukey procedure shows that all three group means differ from each other. Now we can describe the nature of the relationship between caffeine and anagram performance.

- All three groups significantly differed from each other, such that moderate levels of caffeine increased anagram performance (M = 82.00) over no caffeine (M = 69.00), but performance drops off drastically at high levels of caffeine (M = 62.00).

- If only two groups differed, then we would only address this difference in our statement of the nature of the relationship – it was this difference that caused the overall F to be significant.