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# lect5 - IE 410 Notes For Lecture 5 Comparing Individual...

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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 Newman-Keuls 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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lect5 - IE 410 Notes For Lecture 5 Comparing Individual...

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