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Multiple comparisons. - multiple pairwise tests - orthogonal contrasts - independent tests - labelling conventions. Multiple tests. Problem: Because we examine the same data in multiple comparisons, the result of the first comparison affects our expectation of the next comparison.
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Multiple comparisons - multiple pairwise tests - orthogonal contrasts - independent tests - labelling conventions
Multiple tests Problem: Because we examine the same data in multiple comparisons, the result of the first comparison affects our expectation of the next comparison.
3 treatments:-a, no active managementb, selective loggingc, replanting Look at the effect of each treatment on Orang-utan births
significant Not significant Multiple tests ANOVA gives f(2,27) = 5.82, p < 0.05. Therefore at least one different, but which one(s)? Births per km2 • T-tests of all pairwise combinations significant
T-test: <5% chance that this difference was a fluke… affects likelihood of finding a difference in this pair! Multiple tests Births per km2
Multiple tests Solution: Make alpha your overall “experiment-wise” error rate Births per km2 T-test: <5% chance that this difference was a fluke… affects likelihood (alpha) of finding a difference in this pair!
Alpha / 3 = 0.0167 Alpha / 3 = 0.0167 Alpha / 3 = 0.0167 Multiple tests Solution: Make alpha your overall “experiment-wise” error rate e.g. simple Bonferroni: Divide alpha by number of tests Births per km2
Multiple tests Tukey post-hoc testing does this for you Uses the q-distribution Does all the pairwise comparisons. Births per km2 1-2, p < 0.051-3, p > 0.052-3, p < 0.05
Orthogonal contrasts Orthogonal = perpendicular = independent Contrast = comparison Example. We compare the growth of three types of plants: Legumes, graminoids, and asters. These 2 contrasts are orthogonal: 1. Legumes vs. non-legumes (graminoids, asters) 2. Graminoids vs. asters
Trick for determining if contrasts are orthogonal: 1. In the first contrast, label all treatments in one group with “+” and all treatments in the other group with “-” (doesn’t matter which way round). Legumes Graminoids Asters + - -
Trick for determining if contrasts are orthogonal: 1. In the first contrast, label all treatments in one group with “+” and all treatments in the other group with “-” (doesn’t matter which way round). 2. In each group composed of t treatments, put the number 1/t as the coefficient. If treatment not in contrast, give it the value “0”. Legumes Graminoids Asters +1 - 1/2 -1/2
Trick for determining if contrasts are orthogonal: 1. In the first contrast, label all treatments in one group with “+” and all treatments in the other group with “-” (doesn’t matter which way round). 2. In each group composed of t treatments, put the number 1/t as the coefficient. If treatment not in contrast, give it the value “0”. 3. Repeat for all other contrasts. Legumes Graminoids Asters +1 - 1/2 -1/2 0 +1 -1
Legumes Graminoids Asters +1 - 1/2 -1/2 0 +1 -1 0 - 1/2 +1/2 Sum of products = 0 Trick for determining if contrasts are orthogonal: 4. Multiply each column, then sum these products.
Legumes Graminoids Asters +1 - 1/2 -1/2 0 +1 -1 0 - 1/2 +1/2 Sum of products = 0 Trick for determining if contrasts are orthogonal: 4. Multiply each column, then sum these products. 5. If this sum = 0 then the contrasts were orthogonal!
What about these contrasts? 1. Monocots (graminoids) vs. dicots (legumes and asters). 2. Legumes vs. non-legumes
Important! You need to assess orthogonality in each pairwise combination of contrasts. So if 4 contrasts: Contrast 1 and 2, 1 and 3, 1 and 4, 2 and 3, 2 and 4, 3 and 4.
How do you program contrasts in JMP (etc.)? Treatment SS } Contrast 1 } Contrast 2
F1,20 = (67)/1 = 6.7 10 From full model! How do you program contrasts in JMP (etc.)? Legumes vs. non-legumes Normal treatments “There was a significant treatment effect (F…). About 53% of the variation between treatments was due to differences between legumes and non-legumes (F1,20 = 6.7).” Legume 1 1 Legume 1 1 Graminoid 2 2 Graminoid 2 2 Aster 3 2 Aster 3 2 SStreat 122 67 Df treat 2 1 MStreat 60 MSerror 10 Df error 20
Even different statistical tests may not be independent ! Example. We examined effects of fertilizer on growth of dandelions in a pasture using an ANOVA. We then repeated the test for growth of grass in the same plots. Problem?
Multiple tests b Convention: Treatments with a common letter are not significantly different a,b a Births per km2 significant Not significant Not significant