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11 twoway

# 11 twoway - 1 Two-Way Analysis of Variance no interaction...

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1 Two-Way Analysis of Variance - no interaction Example: Tests were conducted to assess the effects of two factors, engine type, and propellant type, on propellant burn rate in fired missiles. Three engine types and four propellant types were tested. Twenty-four missiles were selected from a large production batch. The missiles were randomly split into three groups of size eight. The first group of eight had engine type 1 installed, the second group had engine type 2, and the third group received engine type 3.‘ Each group of eight was randomly divided into four groups of two. The first such group was assigned propellant type 1, the second group was assigned propellant type 2, and so on. Data on burn rate were collected, as follows: Engine Propellant Type type 1 2 3 4 1 34.0 30.1 29.8 29.0 32.7 32.8 26.7 28.9 2 32.0 30.2 28.7 27.6 33.2 29.8 28.1 27.8 3 28.4 27.3 29.7 28.8 29.3 28.9 27.3 29.1 We want to determine whether either factor , engine type (factor A) or propellant type (factor B), has a significant effect on burn rate. Let Y ijk denote the k’th observation at the i’th level of factor A and the j’th level of factor B. The two factor model (without interaction) is: Y ijk = μ + α i + β j + ijk i = 1 , 2 , 3, j = 1 , 2 , 3 , 4, k = 1 , 2, where 1. μ is the overall mean 2. i α i = 0 3. j β j = 0 4. we assume ijk are iid N (0 , σ 2 ) 5. The mean of Y ijk is μ ijk = E [ Y ijk ] = μ + α i + β j This model specifies that a plot of the mean against the levels of factor A consists of parallel lines for each different level of factor B, and a plot of the mean against the levels of factor B consists of parallel lines for each different level of factor A. More generally, there will be I levels of factor A, J levels of factor B, and K replicates at each combination of levels of factors A and B. Y ijk = μ + α i + β j + ijk i = 1 , 2 , . . . , I , j = 1 , 2 , . . . , J , k = 1 , 2 , . . . , K . Here, I = 3, J = 4, K = 2, and there are n = 24 observations in total. There are K = 2 replicates at each level the factors A and B, and the experimental design is said to be balanced , because there are the same number of replicates in each cell. The following table gives the cell means: ¯ Y ij. Engine Propellant Type type 1 2 3 4 1 33.35 31.45 28.25 28.95 2 32.60 30.00 28.40 27.70 3 28.85 28.10 28.50 28.95

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2 Estimated grand mean: ˆ μ = ¯ y ... = 29 . 5917 Estimated factor A level means: ¯ y 1 .. = (33 . 35 + 31 . 45 + 28 . 25 + 29 . 95) / 4 = 30 . 50 ¯ y 2 .. = (32 . 6 + 30 + 28 . 4 + 27 . 7) / 4 = 29 . 675 ¯ y 3 .. = (28 . 85 + 28 . 1 + 28 . 5 + 28 . 95) / 4 = 28 . 60 Estimated factor B level means: ¯ y . 1 . = (33 . 35 + 32 . 6 + 28 . 85) / 3 = 31 . 60 ¯ y . 2 . = (31 . 45 + 30 + 28 . 1) / 3 = 29 . 85 ¯ y . 3 . = (28 . 25 + 28 . 4 + 28 . 5) / 3 = 28 . 383 ¯ y . 4 . = (28 . 95 + 27 . 7 + 28 . 95) / 3 = 28 . 533 Estimation of Model Parameters ˆ μ = ¯ y ... = (30 . 5 + 29 . 675 + 28 . 6) / 3 = (31 . 6 + 29 . 85 + 28 . 383 + 28 . 533) / 4 = 29 . 5917 ˆ α i = ¯ y i.. - ¯ y ... ˆ α 1 = ¯ y 1 .. - ¯ y ... = 30 . 50 - 29 . 5917 = . 908 ˆ α 2 = ¯ y 2 .. - ¯ y ... = 29 . 675 - 29 . 5917 = . 083 ˆ α 3 = ¯ y 3 .. - ¯ y ... = 28 . 60 - 29 . 5917 = - . 992 Note that ˆ α 1 + ˆ α 2 + ˆ α 3 = 0. ˆ β j = ¯ y .j. - ¯ y ... ˆ β 1 = ¯ y . 1 . - ¯ y ... = 31 . 60 - 29 . 5917 = 2 . 0083 ˆ β 2 = ¯ y . 2 . - ¯ y ... = 29 . 85 - 29 . 5917 = . 2583 ˆ β 3 = ¯ y . 3 .
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