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Unformatted text preview: IE 410 Notes For Lecture 5 Comparing Individual treatment Means The general ANOVA test tests: H : all treatment means are equal Suppose H is rejected. We will want to know more about why it was rejected. We have 4 General methods for making comparisons between individual treatment means: 1. Orthogonal Contrasts (in lecture 4) 2. Scheffe's test on all possible contrasts 3. Pairwise comparison tests 4. Comparisons with a control (Dunnett's test) 3. Pairwise comparison tests i) Studentized range methods Tukey's test NewmanKeuls test ii) Duncan's test iii) Least significant difference (LSD) method iv) Fisher significant difference (FSD) or Bonferroni method Variance of a contrast estimator Let L be a general contrast: L = ∑ i c i μ i We will test H : L = 0 Consider an estimate of L, L ˆ L ˆ = ∑ i i i y c We can show that E[ L ˆ ] = L and that V[ L ˆ ] = σ 2 i ∑ c i 2 /n i Recall the following based on our Linear Model : . i y ~ NID( μ i , σ 2 / n i ) ] E( L ˆ ) = E[ i ∑ c i . i y ] = i ∑ E[c i . i y ] = i ∑ c i E[ . i y ] = i ∑ c i μ i V( L ˆ ) =V[ i ∑ c i . i y ] = i ∑ c i 2 V[ . i y ] = i ∑ c i 2 σ 2 /n i Estimating σ 2 with MS E , we can estimate V( L ˆ ) = MS E i ∑ c i 2 /n i This is a key component in most of the methods we will discuss Let us use S( L ˆ ) to denote the standard deviation of the contrast estimate: S( L ˆ ) = i i i E n C MS 2 ∑ Generalized Confidence Interval on a Contrast Let: L ˆ ± CutOff( α )*S( L ˆ ) be a (1 α )*100% CI on the true value of L Recall we are interested in H : L=0 Thus if the CI contains zero, do not reject. This CI is based on our best (unbiased) estimate of L: ( L ˆ ) L ˆ ± (some multiple of)S( L ˆ ) This CI should make intuitive sense....
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 Fall '04
 Storer
 Multiple comparisons, Scheffe

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