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# Experimental Design STAT 507

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This 6 page Class Notes was uploaded by Mr. Alex Berge on Friday October 23, 2015. The Class Notes belongs to STAT 507 at University of Idaho taught by Staff in Fall. Since its upload, it has received 50 views. For similar materials see /class/227942/stat-507-university-of-idaho in Statistics at University of Idaho.

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Date Created: 10/23/15

1 The Scheffe test The F test and criterion used in analysis of variance for a completely randomized design are 55 7 1 F and FD le i The F test and criterion used to test a single contrast7 L7 for a completely randomized design are 7 SSLl F 7 W and FUN i The F test and criterion used for Scheffels test of a single contrast7 L7 for a completely randomized design are 55 g 7 1 F and Fag1y i A comparison of these expressions shows that for Scheffels test the sum of squares for the contrast 55L must exceed the same criterion as SST the entire treatment sum of squares to be declared signi cant This shows why the Scheffe test cannot reject H0 unless the overall ANOVA test is also signi cant7 and it shows why the Scheffe method is conservative 1 Random Effects models Chapter 11 begins with an example of studying variation between cardboard carton machinesi Out of 50 machines 10 are randomly chosen for study We wish to apply the conclusions about these 10 machines to all 50 machines so the chosen groups effects are random effects The model for random effects ANOVA from a completely randomized design looks familiar Yij M l ai l gijv where M is the population grand mean and now the ai terms are random having normal distributions with mean zero and variance 0 The random effects ai are independent of the errors 511quot The null hypothesis of interest in this experiment is H0 a2 0 implying that all ai terms are zero Some other changes with random effects models i We are generally not interested in multiple comparison tests ii It is of interest to test for main effects even in the presence of interactions and iii expected values of mean squares change which for more complicated models can lead to F tests with denominators other than MSEi 11 Constructing F tests As we have previously mentioned the goal of an F test is to arrange it so that under H0 F m l and under HA F gt11 Display llil in our text shows that for the completely randomized design with random effects the usual F MSTREATMSE satis es this criteria Display 112 however shows that for a twofactor random effects model the test for A would be F MSAMSAB in order to satisfy these criteria Display 1113 shows an even stranger situation where there is no term that can be used in the test for A that satis es these criteria We will soon learn how to calculate these expected values of mean squares EMS7s and what to do in situations like in Display 1113 when an exact F test is not available Chapter 11 in our text discusses methods for estimating variance components and con dence intervals for variance components using a method known as the method of moments otherwise known as the ANOVA method We will instead use other methods such as the method of maximum likelihood which is available in SAS Proc MIXEDi 1 More about contrasts and interactions Refer to the computer code for today7s lectures to see how to perform these calculations in SAS 11 Doing 1 way ANOVA as a set of contrasts The ANOVA null hypothesis for 4 groups H0 M1 M2 M3 M4 is equivalent to the following set of contrasts L1 1 1 i 2 0 L2 1 1 i 3 0 L3 1 1 i 4 0 For these three contrasts to be simultaneously equal to zero is equivalent to having all four means equal to each other 12 Contrasts with interactions For an experiment with factorial treatment structure7 with 2 factors each having three levels7 we can look at the treatmentcombination means in the following arrangement B 1 2 3 1 11 M12 13 A 2 21 M22 23 3 31 M32 33 13 Simple main effects tests We can express questions of interest regarding an AxB interaction in the form of contrasts The rst example is simplemain effect tests If we wish to test the equality of factor A when factor B is held at the rst level in our police training example7 we are testing the three patrols when training time is xed at 5 hours we wish to test H0 M11 M21 31 which is equivalent to the following set of contrasts L11M11 M210 L21M11 M310 14 Treatmentcontrast interactions The problem with simplemain effect tests is that they involve a combination of main effects and interaction effects If you repeat the above question for when training time is held at 10 and then 15 hours7 the sum of these three simplemain effect test sum of squares equals SSA SSAB from the factorial model Thus these tests are mixing main effects and interaction effects together An alternative approach is to take a contrast in one factor7 and see if it interacts with the other factor In the police example7 it may be of interest to see if the upperclass patrol middleclass patrol contrast call it 103 M1 7 M2 interacts with the training timer The null hypothesis of no interaction between 1 and training is equivalent to the following set of contrasts L1 1 11 i 21 i 12 i 22 0 L2 1 11 i 21 i 13 i 23 0 If we nd that this interaction is nonsigni cant7 then we may proceed to test the main effect of 1 upperclass patrol middleclass globallyi Several of the ways that the author of our text discusses interaction can addressed in these ways 1 Completely Randomized Design CRD Part I It is the simplest of designs but even planning for it requires scienti c and statistical decisions Acid rain example 11 Exploratory Data Analysis EDA Essential for any statistical analysis For the CRD we often use boxplots for initial examination of the data From the boxplot we can look for evidence of a treatment difference and possible outliers or problems with homogeneity of variance 12 ANOVA as a choice of the best tting model for the mean Let yij l g j l M be the jth observation in group 239 ln ANOVA from a CRD we consider two models for yij The rst model yij Mi 81739 speci es that each group has a different mean value M This is also called the full model for yij The second model yij M 51 speci es that all groups have identical mean values M This is also called the reduced model for 927 Note that the reduced model is a special case or subset of the full model Both models make the assumption that the Eij are independent have mean zero and variance 02 To conduct statistical inference tests con dence intervals etc we make the further assumption that the Eij have a normal distribution An alternative way to express the models above is by letting M Mquot ai where Mquot is the overall mean and ai is the treatment effect of group 239 Then the full model is yij M ai Eij This new formulation of the full model generalizes well to more complicated models but introduces a complication because there are now more parameters than groups For both the full and reduced models we can develop estimators for the parameters M and a2 reduced or M M ai and 02 full shown in Display 31 in the text We can also calculate con dence intervals for our parameters To choose between the full and reduced models we compare their sum of squared residuals SSR A residual 7 is the error in predicting an observation 7 y 7 Q where Qis the predicted value of y For the full model 31739 i the sample group mean and for the reduced model 31739 3 the sample overall mean SSR for the full model can never exceed SSR for the reduced model so we wish to decide if SSR for the full model has been reduced enough to account for the extra parameters in the full model 13 Analysis of Variance mechanics Analysis of variance involves a partition of the total sum of squares for the observations yij Using our notation from above yij 7 y yij 7 E i 7 g which equals the residual from the full model plus the ith treatment effect or Tij 31 By squaring and summing these terms and cancelling the cross product we obtain 9 n g n g n E i if i E2 Z 7 if7 0r SST SSE SSTn i1 j1 i1 j1 i1 j1 As an example consider three groups with the following data Group 1 has ylj values of l 2 and 3 Group 2 has ygj values of 5 3 and 4 and Group 3 has ygj values of 6 7 and 5 The overall sample mean is y 2 231 yij3g 369 4 Then SST is ZEW 7y 7 17 4 2 7 4 5 7 4 7 30 The group means are 31 22 4 and Q3 6 so SSE and SST are 9 n1 SSE 7 22 7a 7 17 2 2 7 2 3 7 2 5 7 4 5 7 6 7 6 and 13971 171 9 7h 9 SST 7 233 7m 7 2mm 7y 7 32 7 4 34 7 4 3lt6 7 4 7 24 171171 71 Thus SST SSE SSTTE or 30 6 24 partitions the total sum of squares about the overall mean into two parts one within groups due to error or effects not accounted by the model and one between groups due to the treatment measuring the difference between sample meansi Since each group here has ni observations each group contributes mi 7 1 degrees of freedom for the error sum of squares for a total of 7 1 degrees of freedom for SSE SST is calculating the sum of squares of 9 sample means about their overall mean so it has 9 7 1 degrees of freedomi For the example data above 7 l gn 71 32 6 and g 7 l 3 l 2 We can summarize this information in an analysis of variance table Source SS df MS F Between groups 24 2 l2 12 Within groups 6 6 1 Total sum of squares 30 8 Three distinct ways to consider these calculations are 1 Focus attention only on the partition of SST 30 6 24 2 To test the null hypothesis H0 M1 M2 3 against the alternative hypothesis Ha some M s differ using an F test or 3 View it as a model selection problem comparing SST SSR for reduced model to SSE SSR for the full model to decide the best tting modeli

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