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# 528 Class Note for STAT 30100 with Professor Dobbin at Purdue

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This 8 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Purdue University taught by a professor in Fall. Since its upload, it has received 14 views.

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Date Created: 02/06/15
Lecture 10 Chapter 13 TwoWay Analysis of Variance Twoway AN OVA compares the means of populations that are classified two ways or the mean responses in twofactor experiments Examples 1 The strength of concrete depends upon the formula used to prepare it An experiment compares six different mixtures Nine specimens of concrete are poured from each mixture Three of these specimens are subjected to 0 cycles of freezing and thawing three are subjected to 100 cycles and three specimens are subjected to 500 cycles The strength of each specimen is then measured Four methods for teaching sign language are to be compared Sixteen students in special education and sixteen students majoring in other areas are the subjects for the study Within each group they are randomly assigned to the methods Scores on a final exam are compared Why is it better to do a TwoWay ANOVA than to just do 2 separate OneWay AN OVAs It is more efficient to study two factors simultaneously rather than separately Your sample size does not have to be as large so experiments with several factors are an efficient use of resources We can reduce the residual variation in a model by including a second factor thought to in uence the response variable lurking variable We are reducing o and increasing the power of the test We can investigate the interactions between factors Assumptions for TwoWay ANOVA l P We have two factors We have I factor levels for the first factor call it factor A and J factor levels for the second factor call it factor B We have I x J combinations of individual factor levels We have independent SRSs of size 111 from each of I x J populations Each of the I x J populations are normally distributed Each of the I x J populations have the same standard deviation 5 Lecture 10 Chapter 13 Page 1 Model for the TwoWay AN OVA Let xyk represent the kth observation from the population having factor A at level I and factor B at level j The statistical model is 2 5 xi c 1 Uk fori l Iandj 1 Jand k 1 my where the deviations Silk are from an N0o distribution Examples For the two examples above identify the response variable and both factors and state the number of levels for each factor I and J and total number of observations N l The strength of concrete depends upon the formula used to prepare it 2 Four methods for teaching sign language are to be compared Sixteen students in special education and sixteen students majoring in other areas are the subjects for the study Lecture 10 Chapter 13 Page 2 General form of the twoway AN OVA table Source Degrees Sum of Mean square F df df pValue of Square Freedom s Factor A DFA SSA MSA SSA FA MSA Main Il DFA MSE effect of DFA DFE A Factor B DFB SSB MSE SSB FEMSB Main Jl DFB MSE effect of DFB DFE B A amp B DFAB SSAB MSAB SSAB MSAB Interactio 7 AB 1nteraction I l J l DFAB MSE n of A DFAB DFE and B Error DFE SSE MSE SSE Within N U DFE Total N l SST MST SST DF T The three hypotheses that are tested are H 0 I main effect of Factor A is zero H 0 I main effect of Factor B is zero H 0 I interaction between A and B is zero Doing the ANOVA is not sufficient Also look at graphs of the marginal means of the combinations to interpret the results Graphical example of no effect effect ofA effect ofB Interaction between A and B Lecture 10 Chapter 13 Page 3 Example Exercise 1314 One way to repair serious wounds is to insert some material as a scaffold for the body s repair cells to use as a template for new tissue Scaffolds made from extracellular material ECM are particularly promising for this purpose Because they are made from biological material they serve as an effective scaffold and are then reabsorbed One study compared 6 types of scaffold material Three of these ECMs and the other three were made of inert materials There were 3 mice used per scaffold type The response measure was the of glucose phosphated isomerase Gpi cells in the region of the wound A large value is good indicating that there are many bone marrow cells sent by the body to repair the tissue Here are the data for 2 weeks 4 weeks and 8 weeks after the repair Material Gpi 2 weeks 4 weeks 8 weeks ECM1 7O 55 6O 75 7O 65 65 7O 65 ECM2 6O 6O 6O 65 65 7O 7O 65 6O ECM3 80 75 7O 6O 7O 80 75 75 7O MAT1 5O 20 1 5 45 25 25 5O 25 25 MAT2 5 5 1 O 1 O 1 O 5 1 5 5 5 MAT3 3O 1 O 5 25 1 5 1 5 25 1 O 1 0 Lecture 10 Chapter 13 Page 4 a Make a table giving the sample size mean and standard deviation for each of the materialbytime combinations Is it reasonable to pool the variances Because the sample sizes in this experiment are very small we expect a large amount of variability in the sample standard deviations Although they vary more than we would prefer we will proceed with the AN OVA SPSS DataSplit le select compare groups move one categorical variable material into groups box Analyzecompare means VIeans move the other categorical variable weeks into independent list move gpi into dependent list Descriptive Statistics Dependent Variable gpi material I weeks Mean Std Deviation N ECM1 2 Weeks 7000 5000 3 4 Weeks 6500 8660 3 8 weeks 6333 2887 3 Total 6611 6009 9 ECM2 2 weeks 6500 5000 3 4 weeks 6333 2887 3 8 weeks 6333 5774 3 Total 6389 4167 9 ECM3 2 Weeks 7167 10408 3 4 weeks 7333 2887 3 8 weeks 7333 5774 3 Total 7278 6180 9 MAT1 2 weeks 4833 2887 3 4 weeks 2333 2887 3 8 weeks 2167 5774 3 Total 3111 13411 9 MAT2 2 weeks 1000 5000 3 4 weeks 667 2887 3 8 weeks 667 2887 3 Total 778 3632 9 MAT3 2 weeks 2667 2887 3 4 weeks 1167 2887 3 8 weeks 1000 5000 3 Total 1611 8580 9 Total 2 weeks 4861 24363 18 4 weeks 4056 28330 18 8 weeks 3972 28619 18 Total 4296 26961 54 Lecture 10 Chapter 13 Page 5 b Make a table giving the sample size mean and standard deviation for each of the material types Give a short summary of the Gpi depends on material SPSS DataSplit le remove the compare groups Analyzecompare means VIeans move material into independent list move gpi into dependent list Repott gpi material Mean N Std Deviation ECM1 6611 9 6009 ECM2 63 89 9 4167 ECM3 7278 9 6180 MAT1 3111 9 13411 MAT2 778 9 3632 MAT3 1611 9 8580 Total 42 96 54 26 961 3 Make a table of the sample size mean and standard deviation for the weeks after the repair Give a short summary of the Gpi depends on the weeks after repair SPSS same only move weeks into independent box Repott gpi weeks Mean N Std Deviation 2 weeks 4861 18 24363 4 weeks 4056 18 28330 8 weeks 3972 18 28619 Total 4296 54 26 961 Lecture 10 Chapter 13 Page 6 d Run the analysis of variance Report the F statistics with degrees of freedom and P Values for the main effects and the interaction What are the hypotheses you are testing What do you conclude SPSS Instructions AnalyzegtGeneral LineargtUnivariate Move the appropriate variables to the dependent and xed factor boxes Select Model and do not remove the checkmark beside the Include intercept in model Tests of BetweenSu bjects Effects Dependent Variable gpi Type III Sum Source of Squares df Mean Square F Sig Corrected Model 37609259a 17 2212309 86883 000 Intercept 99674074 1 99674074 3914473 000 material 35659259 5 7131852 280087 000 weeks 867593 2 433796 17036 000 material weeks 1082407 10 108241 4251 001 Error 916667 36 25463 Total 138200000 54 Corrected Total 38525325 53 a R Squared 976 Adjusted R Squared 965 Lecture 10 Chapter 13 Page 7 e Make a plot of the means of the combinations Describe the main features of the plot SPSS Instructions Again AnalyzeGeneral LinearUnivariate Move the appropriate variables to the dependent and xed factor boxes To do plots select the Plots box and move factor variables to appropriate horizontal axis or lines boxes click add then click continue then click OK Lecture 10 Chapter 13 Page 8

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