22ANOVA_Intro

22ANOVA_Intro - Statistical Techniques I EXST7005...

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Statistical Techniques I EXST7005 Conceptual Intro to ANOVA
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Analysis of Variance (ANOVA) R. A. Fisher - resolved a problem that had existed for some time. H0: μ 1 = μ 2 = μ 3 = . .. = μ k H1: some μ i is different Conceptually, we have separate (and independent) samples, each giving a mean, an we want to know if they could have come from the same population or if is more likely they come from different populations.
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The Problem (continued) One way to do this is a series of t-tests. If we want to test among 3 means we do 3 tests: 1 versus 2, 1 versus 3, 2 versus 3 For 4 means there are 6 tests. 1-2, 1-3, 1-4, 2-3, 4, and 3-4 For 5 means, 10 tests, etc.
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The Problem (continued) This technique is unwieldy, and worse. When we do the first test, there is an α chance error, and for each additional test another α chance of error. So if you do 3 or 6 or 10 tests, the chance of error on each and every test is α . Overall, for the experiment, the chance of error for all tests together is much higher than α .
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The Problem (continued) Bonferroni gave a formula that showed that the chance of error would be NO MORE than Σα i. S if we do 3 tests, each with a 5% chance of error the overall probability of error is no greater than 15%, 30 percent for 6 tests, 50% for 10 tests, etc. Of course this is a lower bound. A better calculation comes from a the value. α '=1-(1- α )k
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No. of means pairwise tests Bonferroni's lower bound Duncan's 1-(1- α )k (1- α ) 2 1 0.05 0.0500 0.950 3 3 0.15 0.1426 0.857 4 6 0.30 0.2649 0.735 5 10 0.50 0.4013 0.599 6 15 0.75 0.5367 0.463 7 21 1.05 0.6594 0.341 10 45 2.20 0.9006 0.099 50 1225 61.20 0.9999 0.000 100 4950 247.45 1.0000 0.000 for α = 0.05
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(continued) The bottom line: Splitting an experiment into a number of smaller tests is generally a poor idea This applies at higher levels as well (i.e. splitting big ANOVAs into little ones). The solution: We need ONE test that will give u
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22ANOVA_Intro - Statistical Techniques I EXST7005...

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