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Lecture 15

ADM2304 Lecture 15: 15
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Department
Administration
Course
ADM2304
Professor
Tony Quon
Semester
Fall

Description
Main Effect of Factor B Hypotheses: H0: No effect of Factor B (that is average change time independent of vehicle) or j = 0 for all j Ha: There is an effect due to Factor B or j ≠ 0 for some j If Factor B has little effect then we expect the mean of each level to be close to the overall mean and therefore the F statistic below should be small 2 ∑ a×r× x(−•j ) j F = MSB = b−1 ~ F MSE MSE b−1,ab(r−1) Example: Brake Replacement Source Degrees Sum oMean SF-statistic VariatiFreedoSquareof Squares -> Compare with earlier table that included the interaction. Manufacture2 595.8 297.9 7.73 First two rows and the last row are the same but now the Vehicle 1 6840.36840.3177.46 Total 29 8438.338.55 “interaction” sum of squares has been included in the error sum of squares Multiple Pairwise Comparisons —> Comparing the means allows us to explore the nature of the differences detected by the F-tests When there is interaction between the factors in how they affect the response variable, the main interest is comparing the treatment means When there are main effects, then we may be interested in comparing row means or column means, or comparing Factor A or Factor B means Comparison of Treatment Means —> For each set of comparisons, we start by determining how many means there are to compare: In our example, the primary concern is with the pairwise comparisons of the ab = 6 treatment means There are a total of J = 6×5/2 = 15 pairwise comparisons Because each comparison is effectively a hypothesis test, the multiple comparisons demands an adjustment of the level of significance for each individual test to ensure that the overall level of significance is no larger than Comparing Treatment Means —> To compare each pair of treatment means, we look at the observed differences and compare them to the ± margin of error: ⎛ ⎞ ⎜ ⎛1 1 ⎞ ⎟ ⎜xij• ±tkl•SE ⎜ + ⎟ ⎟ ⎝ ⎝ r r⎠ ⎠ —> where t* is the upper α/(2J) point of the t-distribution with degrees of freedom given by the MSE (ab(r - 1)) and r is the number of observations that comprise each treatment mean Calculations Here, if α = .05, then α/(2J) = .05/ (2×15) = .05/30 = .001667 Since the MSE has 24 d.f., the critical value of t is 3.2583: Student's t distribution with 24 DF P( X <= x ) x 0.998333 3.25830 The margin of error is: 3.2583×sqrt(34.53×2/5) = 3.2583×3.71645 = 12.109 The six treatment means are: (A, Car) = 60.2, (B, Car) = 65, (C, Car) = 74.6 (A, SUV) = 97.2, (B, SUV) = 91.8, (C,SUV) = 101.4 Calculations By inspection: The pairs (60.2,65), (65
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