Lec 1: An Introduction to ANOVA
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1 Ying Li Stockholm University October 31, 2011
2 Three end-aisle displays Which is the best?
3 Design of the Experiment Identify the stores of the similar size and type. The displays are randomly assigned to use.
4 Design of the Experiment Identify the stores of the similar size and type. The displays are randomly assigned to use. First Principle Randomization
5 Observations 3 levels 5 replicates
6
7 The Analysis of Variance a level of the factors (a treatments) n replicates N = a n runs Completely randomized design
8 Statistical Model y ij = µ + τ i + ε ij i = 1, a j = 1, n µ : overall mean τ i : the effect of ith treatment ε ij N(0, σ 2 ),i.i.d. a i τ i = 0
9 The Analysis of Variance We are interested in testing the equality of treatment means E(ȳ i. ) = µ + τ i = µ i i = 1, a The appropriate hypothesis are H 0 : µ 1 = µ 2 = = µ a H 1 : µ i µ j for at least one pair (i, j)
10 If H 0 is true, the distribution of F 0 is F a 1,a(n 1) Reject H 0 if F 0 > F α,a 1,a(n 1)
11
12 Equation for manual calculation Balanced Data: a n SS T = i=1 j=1 ij y.. 2 N y 2 Unbalanced Data SS T = a n i i=1 j=1 ij y.. 2 N y 2 SS Treatment = 1 n a i=1 i. y.. 2 N y 2 SS Treatment = a yi. 2 y.. 2 n i N i=1 SS E = SS T SS Treatment SS E = SS T SS Treatment
13 Example ȳ 1. = 5.73,ȳ 2. = 6.24, ȳ 3. = 8.32,ȳ.. = 6.76 SS T = SS Treatment = SS E = SS T SS Treatment = 3.15
14 DF SS MS F P Treatment <0.001 Error Total
15 DF SS MS F P Treatment <0.001 Error Total Which ones cause the differ?
16 Multiple Comparison Methods We know the end-aisle display different than others. We might suspect the first and the second are different. Thus one reasonable test hypothesis would be H 0 : µ 1 = µ 2 H 1 : µ 1 µ 2 or, H 0 : µ 1 µ 2 = 0 H 1 : µ 1 µ 2 0
17 Contrasts In general, a contrast is a linear combination of the parameters of the form a Γ = c i µ i i=1 where c i are the contrast constants, and a i=1 c i = 0. For unbalance design a i=1 c in i = 0.
18 The hypothesis can be expressed in the terms of contrasts H 0 : a c i µ i = 0 i=1 a H 1 : c i µ i 0 i=1 In our example, c 1 = 1,c 2 = 1, c 3 = 0.
19 Contrast Tests Testing hypothesis involving contrast can be done in two basic ways 1 t-test: 2 F-test: where a i=1 t 0 = c iȳ i. MS E a n i=1 c2 i F 0 = MS c MS E = SS c/1 MS E SS c = ( a i=1 c iȳ i. ) 2 1 n a i=1 c2 i
20 Orthogonal Contrasts Two contrasts a i=1 = c iµ i = 0 and a i=1 = d iµ i = 0 are orthogonal if a c i d i = 0 or in unbalanced case if i=1 a n i c i d i = 0 i=1 We can specify a 1 orthogonal contrasts. Tests performed on orthogonal contrasts are independent.
21 Example
22 Pairwise Comparison The null hypothesis are H 0 : µ i = µ j, i j
23 Pairwise Comparison The null hypothesis are H 0 : µ i = µ j, i j Tukey s test Make use of the distribution of studentized range statistics q = ȳmax ȳ min MSE 2 ( 1 n i + 1 n j ) Appendix Table 7 contains upper percentiles for q. The difference d between two averages is significant if d > q α(a, f ) 2 MS E ( 1 n i + 1 n j )
24 The fisher least significant difference method (LSD) The fisher least significant difference method procedure use T test for testing H 0 : µ i = µ j, i j The test statistics is: ȳ i. ȳ j. t 0 = MS E ( 1 n i + 1 n j ) The quantity: least significant difference (LSD) If LSD = t α/2,n a MS E ( 1 n i + 1 n j ) ȳ i. ȳ j. > LSD we conclude that the population means µ i and µ j differ.
25 Questions:1 Why in some situation overall F test of ANOVA is significant, but the pairwise comparison fails to reveal the differences?
26 Questions:2 Many tests available: Duncan, Student-Newman-Keuls, REGWQ, Bonferroni, Sidak, Scheff, Which pairwise comparison method do I use?
27 Questions:2 Many tests available: Duncan, Student-Newman-Keuls, REGWQ, Bonferroni, Sidak, Scheff, Which pairwise comparison method do I use? unfortunately, no-clear cut answer. Fisher s LSD procedure, only apply if F test is significant Tukey s method control overall error rate, many statisticians prefer.
28 Dunnett s Test Often one of the treatments is a control treatment, and we want to compare the other treatments with the control treatment. We want to test a 1 hypotheses: H 0 : µ j = µ 1 where i = 2,, a
29 Dunnett s Test For the ith treatment, H 0 is rejected if ȳ i ȳ 1 > d α (a 1, f ) MS E ( ) n i n j Appendix 8 gives upper percentiles for d α (a 1, f )
30 One Example
31 One Example
32 One Example
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