# UNIT2 - Unit 2 Experiments with a Single Factor One Way...

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Unformatted text preview: Unit 2: Experiments with a Single Factor: One Way ANOVA Sources : Sections 1.6 to 1.9, additional materials (in these notes) on random effects . One-way layout with fixed effects. Multiple comparisons. Quantitative factors and orthogonal polynomials. Residual analysis. One-way layout with random effects. 1 One-way layout and ANOVA: An Example Reflectance data in pulp experiment: each of four operators made five pulp sheets; reflectance was read for each sheet using a brightness tester. Randomization : assignment of 20 containers of pulp to operators and order of reading. Table 1: Reflectance Data, Pulp Experiment Operator A B C D 59.8 59.8 60.7 61.0 60.0 60.2 60.7 60.8 60.8 60.4 60.5 60.6 60.8 59.9 60.9 60.5 59.8 60.0 60.3 60.5 Objective : determine if there are differences among operators in making sheets and reading brightness. 2 Model and ANOVA Model : y i j = + i + i j , i = 1 ,..., k ; j = 1 ,..., n i , where y i j = j th observation with treatment i , i = i th treatment effect, i j = error, independent N ( , 2 ) . Model fit: y i j = + i + r i j = y +( y i y )+( y i j y i ) , where . means average over the particular subscript. ANOVA Decomposition : k i = 1 n i j = 1 ( y i j y ) 2 = k i = 1 n i ( y i y ) 2 + k i = 1 n i j = 1 ( y i j y i ) 2 . 3 F-Test ANOVA Table Degrees of Sum of Mean Source Freedom ( d f ) Squares Squares treatment k 1 SST = k i = 1 n i ( y i y ) 2 MST = SST / d f residual N k SSE = k i = 1 n i j = 1 ( y i j y i ) 2 MSE = SSE / d f total N 1 k i = 1 n i j = 1 ( y i j y ) 2 The F statistic for the null hypothesis that there is no difference between the treatments, i.e., H : 1 = = k , is F = k i = 1 n i ( y i y ) 2 / ( k 1 ) k i = 1 n i j = 1 ( y i j y i ) 2 / ( N k ) = MST MSE , which has an F distribution with parameters k 1 and N k . 4 ANOVA for Pulp Experiment Degrees of Sum of Mean Source Freedom ( d f ) Squares Squares F operator 3 1.34 0.447 4.20 residual 16 1.70 0.106 total 19 3.04 Prob ( F 3 , 16 &amp;gt; 4 . 20 ) = 0.02 = p value, thus declaring a significant operator-to-operator difference at level 0.02. Further question: among the 6 = ( 4 2 ) pairs of operators, what pairs show significant difference? Answer: Need to use multiple comparisons. 5 Multiple Comparisons For one pair of treatments, it is common to use the t test and the t statistic t i j = y j y i 1 / n j + 1 / n i , where n i = number of observations for treatment i , 2 = RSS/df in ANOVA; declare treatments i and j different at level if | t i j | &amp;gt; t N k , / 2 ....
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UNIT2 - Unit 2 Experiments with a Single Factor One Way...

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