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onefactors2010 - Statistics 514 Experiments with One Single...

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Statistics 514: Experiments with One Single Factors: ANOVA Lecture 3. Experiments with a Single Factor: ANOVA Montgomery 3-1 through 3-3 Page 1
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Statistics 514: Experiments with One Single Factors: ANOVA Tensile Strength Experiment Investigate the tensile strength of a new synthetic fiber. The factor is the weight percent of cotton used in the blend of the materials for the fiber and it has five levels. percent tensile strength of cotton 1 2 3 4 5 total average 15 7 7 11 15 9 49 9.8 20 12 17 12 18 18 77 15.4 25 14 18 18 19 19 88 17.6 30 19 25 22 19 23 108 21.6 35 7 10 11 15 11 54 10.8 Page 2
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Statistics 514: Experiments with One Single Factors: ANOVA Data Layout for Single-Factor Experiments treatment observations totals averages 1 y 11 y 12 · · · y 1 n y 1 . ¯ y 1 . 2 y 21 y 22 · · · y 2 n y 2 . ¯ y 2 . . . . . . . . . . · · · . . . . . . . . . a y a 1 y a 2 · · · y an y a. ¯ y a. Page 3
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Statistics 514: Experiments with One Single Factors: ANOVA Analysis of Variance Interested in comparing several treatments. Could do numerous two-sample t-tests but this approach does not test equality of all treatments simultaneously. ANOVA provides a method of joint inference. Statistical Model: y ij = μ + τ i + ij i = 1 , 2 . . . a j = 1 , 2 , . . . n i μ - grand mean; τ i - i th treatment effect; ij N (0 , σ 2 ) - error Constraint: a i =1 τ i = 0 . Basic Hypotheses: H 0 : τ 1 = τ 2 = . . . = τ a = 0 vs H 1 : τ i 6 = 0 for at least one i Page 4
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Statistics 514: Experiments with One Single Factors: ANOVA Derive Estimates: Partitioning y ij Notation y i. = n i j =1 y ij y i. = y i. /n i (treatment sample mean, or row mean) y .. = ∑ ∑ y ij y .. = y .. /N (grand sample mean) Decomposition of y ij : y ij = y .. + ( y i. - y .. ) + ( y ij - y i. ) Estimates for parameters: ˆ μ = y .. ˆ τ i = ( y i. - y .. ) ˆ ij = y ij - y i. ( residual ) So y ij = ˆ μ + ˆ τ i + ˆ ij . It can be verified that a i =1 n i ˆ τ i = 0; n i j =1 ˆ ij = 0 for all i. Page 5
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Statistics 514: Experiments with One Single Factors: ANOVA Test Basic Hypotheses: Partitioning the Sum of Squares Recall y ij - y .. = ˆ τ i + ˆ ij = ( y i. - y .. ) + ( y ij - y i. ) Can show i j ( y ij - y .. ) 2 = i n i ( y i. - y .. ) 2 + i j ( y ij - y i. ) 2 = i n i ˆ τ 2 i + i j ˆ 2 ij .
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