Statistical Techniques IIEXST7015Analysis of Variance15a_ANOVA_Introduction 1DesignThe simplest model for Analysis of Variance (ANOVA) is the CRD, the Completely Randomized DesignThis model is also known a "One-way" Analysis of Variance. Unlike regression, which fits slopes for regression lines and calculates a measure of random variation about those lines, ANOVA fits means and variation about those means. 15a_ANOVA_Introduction 2Design (continued)The hypotheses tested are hypotheses about the equality of meansH0: µ1= µ2= µ3= µ4= ... = µt Where the µirepresent means of the levels of some categorical variable"t" is the number of levels in the categorical variable. H1: some µiis different 15a_ANOVA_Introduction 3Design (continued)We will generically refer to the categorical variable as the "treatment" even though it may not actually be an experimenter manipulated effect. The number of treatments will be designated "t". The number of observations within treatments will be designated n for a balanced design (the same number of observations in each treatment), or nifor an unbalanced design (for i = 1 to t). 15a_ANOVA_Introduction 4Design (continued)The assumptions for basic ANOVA are very similar to those of regression. The residuals, or deviations of observations within groups should be normally distributed. The treatments are independently sampled. The variance of each treatment is the same (homogeneous variance). 15a_ANOVA_Introduction 5ANOVA reviewI am borrowing some material from my EXST7005 notes on t-test and ANOVA. See those notes for a more complete review of the introduction to Analysis of Variance (ANOVA). Start with the logic behind ANOVA. 15a_ANOVA_Introduction 6ANOVA review (continued)Prior to R. A. Fisher's development of ANOVA, investigators were likely to have used a series of t tests to test among t treatment levels. What is wrong with that? Recall the Bonferroni adjustment. Each time we do a test we increase the chance of error. To test among 3 treatments we need to do 3 tests, among 4 treatments, 6 tests, 5 treatments are 10 tests, etc. 15a_ANOVA_Introduction 7ANOVA review (continued)What is needed is ONE test for a difference among all tests with one overall value of αspecified by the investigator (usually 0.05). Fisher's solution was simple, but elegant. 15a_ANOVA_Introduction 8ANOVA review (continued)Suppose we have a treatment with 5 categories or levels. We can calculate a mean and variance for each treatment level. In order to get one really good estimate of variance we can pool the individual variances of the 5 categories (assuming homogeneity of variance). This pooled variance can be calculated as a weighted mean of the variance (weighted by the degrees of freedom). 15a_ANOVA_Introduction 9ANOVA review (continued)YGroupABCDEY 15a_ANOVA_Introduction 10ANOVA review (continued)And since (n1-1)S12= SS1, the weighted mean is simply the sum of the SS divided by the sum of the d.f. SSS SS SS SS SSnnnnnp1234512345211111=++++−+ −+ −+ −+ −()()()()()SnSnSnSnSnSnnnnnp1122223324425521234521111111111=−+−+−+−+−−+ −+ −+ −+ −()()()()()()()()()()15a_ANOVA_Introduction 11ANOVA review (continued) 15a_ANOVA_Introduction 12ANOVA review (continued)So we have one very good estimate of the random variation, or sampling error, S2. Then what? 15a_ANOVA_Introduction 13ANOVA review (continued)Now consider the treatments. Why don't they all fall on the overall mean? Actually, under the null hypothesis, they should, except for some random variation. So if we estimate that random variation, it should be equal to the same error we already estimated within groups? 15a_ANOVA_Introduction 14ANOVA review (continued)If we estimate a variance with means, we are estimating the variance of means, which is S2/n. If we multiply this by "n" it should actually be equal to S2, which we estimated with S2p, the pooled variance estimate. 15a_ANOVA_Introduction 15ANOVA review (continued)So if the null hypothesis is true, the mean square of the deviations within groups should be equal to the mean square of the deviations of the means multiplied by "n"!!!!15a_ANOVA_Introduction 16ANOVA review (continued) YGroupABCDEYDeviationsbetweengroupsMeansDeviationswithingroups15a_ANOVA_Introduction 17ANOVA review (continued)Now, if the null hypothesis is not true, and some µiis different, then what? YGroupABCDEY15a_ANOVA_Introduction 18ANOVA review (continued)Then, when we calculate a mean square of deviations of the means from the overall mean, it should be larger than the previously estimated S2p. So we have two estimates of variance, S2p and the variance from the treatment means. If the null hypothesis is true, they should not be significantly different. 15a_ANOVA_Introduction 19ANOVA review (continued)If the null hypothesis is FALSE, the treatment mean square should be larger. It will therefore be a ONE TAILED TEST!15a_ANOVA_Introduction 20ANOVA review (continued)We usually present this in an "Analysis of Variance" table. Source d.f.Sum of SquaresMean SquareTreatment t-1 SSTreatment MSTreatmentError t(n-1) SSError MSErrorTotal tn-1 SSTotal15a_ANOVA_Introduction 21ANOVA review (continued)Degrees of freedomThere are tn observations total (Σniif unbalanced). After the correction factor, there are tn-1 for the corrected total.There are t-1 degrees of freedom for the t treatment levels. Each group contributes n-1 d.f. to the pooled error term. There are t groups, so the pooled error (MSE) has t(n-1) d.f.15a_ANOVA_Introduction 22ANOVA review (continued)The SSTreatments is the SS deviations of the treatment means from the overall mean. Each deviation is denoted τi, and is called an treatment "effect". SSTreatments = (Y - Y)i2i=1tii=1t∑∑=τ215a_ANOVA_Introduction 23ANOVA review (continued)The model for regression isYi= β0+ β1Xi+ εi The effects model for a CRD is Yij= µ + τi+ εij where the treatments are i=1, 2, ... tand the observations are j=1, 2, ... n, or ni for unbalanced data The means model is Yij= µi+ εij 15a_ANOVA_Introduction 24ANOVA review (continued)UncorrectedSSTreatments = Yijj=1ni=1t()∑∑2The calculations. The SSTotal is exactly the same as regression, the sum of all ΣY2ij observations (squared first). The correction factor is exactly the same too, all observations are summed, the sum is squared and divided by the number of observations, (ΣYij)2/tn. 15a_ANOVA_Introduction 25ANOVA review