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Discussion 31/20/2014OutlineHow so that fill out the table in the appendix
Central Michigan University, US has reference to this Academic Journal, Discussion 31/20/2014OutlineHow so that fill out the table in the appendix in HW3What does the Model statement do in SAS Proc GLM (please download lab 2 in consideration of reference) What is a statistical model in layman?s termsWhat are the residuals in addition to predicted valuesHow so that calculate Variance ComponentsPower questionsQuestions?
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How so that fill out the table in the appendix in HW3The treatments are the herbicides used in addition to there are 6 different herbicides (treatments)The pots are the objects that were applied the treatment therefore they are your experimental unitEach pot contained 4 plants therefore 4 measurements were made so that each pot (sub-samples)How so that fill out the table in the appendix in HW3What does the Model statement do in SAS Proc GLMWhat is a statistical model in layman?s termsA mathematical equation constructed so that describe the makeup of an observation or a group of observationsExample: One-way ANOVA modelThe One-way ANOVA model describes an observation (Y) as a deviation from an overall mean (æ) of a group of observations due so that a treatment effect (?) in addition to the addition of random error (?).
What does the Model statement do in SAS Proc GLMThe overall mean (?.) is the mean of all the observations Yij Yij read as observation from the ith treatment in addition to the jth replication.The treatment effect is the deviation from the overall mean ?. so that the treatment mean (?i.)The random error is the deviation from the treatment mean (?i.) so that a given replication of that treatment (Yij )What are the predicted values?The theoretical values obtained based on the statistical model (the error is excluded from the model):What are the residuals?The deviation from the expected values so that the observed values:What does the Model statement do in SAS Proc GLMPredicted Yij( ) To do ANOVA in SAS we use Proc GLM in addition to specify our model:Example 3.2 in lab 2:What does the Model statement do in SAS Proc GLMProc GLM; Class Culture; Model Nlevel = Culture; Means Culture; Output Out = Residual R = Res1 P = Pred1; The Class statement tells SAS that our data is grouped by the variable Culture in the model statement we tell SAS that we want so that explain Nlevel by the variable Culture; therefore:SAS calculates the overall mean in addition to the residuals but what Proc GLM is also calculating is the sums of squares, mean squares, F ? values, in addition to p ?values in consideration of each variable we specify in the model see how in the following slide.
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How does SAS calculate the sums of square (SS)?What does the Model statement do in SAS Proc GLMWhich is equivalent to:When you divide the SS by their respective degrees of freedom the mean squares are obtained (Equivalent so that the variances, s2)Where: r = number of replicationsIn Nested Designs:Example 3.4 in lab 2:What does the Model statement do in SAS Proc GLMProc GLM; Class Trtmt Pot; * We want SAS so that calculate the variances between pots because that will be our error in consideration of our ANOVA Model Growth = Trtmt Pot(Trtmt); *Pot is not a treatment. Pot is only an ID variable Random Pot(Trtmt); *must specify pot as random because we are not interested in detecting differences between pots Test H = Trtmt E = Pot(Trtmt); * Here we request a customized F testWhere i is treatment ID, j is replication ID, k is subsample IDr = number of replications, s = number of sub-samplesTotal SSTreatment SSPot SS (e.u.)sub-sample SSHow so that Calculate Variance ComponentsWe analyze nested designs so that estimate the variance components which can be used so that estimate optimal sub-sample size (section 220.127.116.11 in lecture reading topic 3)The variance components are the estimate of variance in consideration of a particular variable (e.g. treatment, experimental unit, in addition to subsample)The variance components can be calculate using Proc VarComp in SAS
In the lecture topic 3 reading section 18.104.22.168 an experiment is described where mint plants are exposed so that different treatments of temperature in addition to daylight in addition to stem length was measuredThere are a total of 6 treatments, 3 pots (replications) per treatment, in addition to 4 plants were measured per pot (subsamples)How so that Calculate Variance ComponentsThe sums of squares was calculated in consideration of treatment, pots in addition to subsamplesThen the variances (equivalent so that means squares)How so that Calculate Variance ComponentsTotal SSTreatment SSPot SS (e.u.)sub-sample SSMS Treatment MS PotMS sub-samplet -1t (r -1)rt (s – 1)Where t = number of treatments, r = number of replications, in addition to s = number of sub-samplesThe variance due so that the sub-sample is the variance due so that error:How so that Calculate Variance ComponentsMS sub-sample = ??2To estimate the variance component of the subsample we just calculate the MS of sub-sample=0.93
The variance of pots contains the variance of the sub-samples (NOTE: this is not the variance component of pots):How so that Calculate Variance ComponentsMS Pots = ??2 + 4??2To estimate the variance component of pots we have so that solve in consideration of ??2. The variance components of pot is calculated below:??2 = (MS Pots – ??2) / 4??2 = (2.15 – 0.93) / 4 = 0.30= 2.15The variance of treatments includes the variance of pots in addition to subsamples (NOTE:):How so that Calculate Variance ComponentsMS Treatments = ??2 + 4??2 + 12??2/5 = 35.92To estimate the variance component of treatments only we have so that solve in consideration of ??2/5??2/5 = (MS Treatments – ??2 – 4??2) / 12 ??2/5 = (35.92 – 0.93 ? 4*0.3) / 12 = 2.81 PowerHow so that use Power Charts in consideration of ANOVA
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