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# 118 Class Note for STAT 401 at PSU

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Date Created: 02/06/15

STAT 401 Ch 10 OneWay Analysis of Variance ANOVA In this chapter we consider a method for the comparison the mean of 1 populations and the related issue of multiple com parisons New Terminology It is common to refer to the 1 populations as I factorlevels EG if we want to compare the mean yield of a chemical process under I 4 different temperature settings the factor is temperature and the setting each of which de nes a population are the levels Let uni 1 1 denote the mean of the i th population or factor level Of interest is testing H0 1 2 1 vs Ha H0 is false Data The data from the ith factor level are XihXiz XU so sample sizeJ for all 1 Note We will cover only the case where the sample sizes are the same Assumptions A111 populations are normal with the same vari ance That is XUN cszforaiii11j11 Model This is the simplest situation for which it is common to write a statistical model for the data Asz39jui8ij l I where y FEM t y t and 8 N N062 The differ ence on y y of y from the average mean value y is called the effect of the ith population or factor level of treatment The null hypothesis H0 1 2 y can be written equiv alently asHo X1 X2 2090 Question Since we can test the equality of two means why don t we test H0 by testing equality of all pairwise means Basic Idea for Analysis of Variance Suppose I 3 and 1 lt 2 lt 3 The three samples would tend to look like J 6 Sample 1 Sample 2 Sample 3 Combined S On the other hand if m 2 3 then the picture might look like Sample 1 Sample 2 Sample 3 Combined S Note that the variability in the combined sample is larger in the case that m lt 2 lt 3 This is because the combined sample variability is due to a the variability within each sample and b the variability between Exit 1 Notation 1 X j i th sample mean l 1 J 11 X Z ZXijzgrand mean ell2111 l J 111 2 Xij X2 ith sample variance 121 To turn the above basic idea for analysis of variance into a test procedure we argue as follows 1 The common variance 62 can be estimated by pooling the m I JFrom rules of expectation it follows that MSE Emmmamp regardless of the values of 1 2 y 3 2 2 Since VarX 671 1 I it follows that when H0 1 2 y is true the sample variance offh 9 will be an unbiased estimator of 621 or MSTr J XL 43 is an unbiased estimator of 62 Thus EMSTr 62 ifHO is true However EMSTr gt 62 ifHO is false The test procedure is based on a comparison of the two variance estimators MSTr MSE If H0 is true and the assumptions are met TEst statistics F F NFlIJUl ie it has an F distribution with I l and I J 1 degrees of freedom Rejection rule at level or F gt 714104 The critical values for the F distr are in Table A7 4 Computational Formulas We will use initials SST Total Sum of Squares SSTr Treatment Sum of Squares SSE Error Sum of Squares and the additional notation I J J i1j1 11 SST ZZXJ X2zzgggli 3 i j l j 2 l 2 2 SSTr zg Xz JXL LIX SSE 220g X2 SST SSTr 139 j Note The identity SST SST r SSE is responsible for the name Analysis of Variance SSTrI 1 T F est Statistic SSE1J 1 It is common to denote SSTr 1 1 SSE MSE 1J 1 which equal the quantities we saw before Thus MST r 2 Mean SS due to Treatment 2 Mean SS due to Error MSTr T tSt tit F es a1s1c MSE Ex1 The following data resulted from comparing the degree of soiling in fabric treated with three different mixtures of methacrylic acid MiXl 56 112 90 107 94 X1459 2E1918 MiX2 72 69 87 78 91 X2397 X2794 MiX3 62 108 107 99 93 X3 469 X3 938 TestHo 11 2 03 vs Ha H0 is false at 01 01 S01 Firstziszi j 5621122 1 1 932 121351 221 X1 X2 X3 1325 Thus 1 SST 121351 13252 4309 015 1 SSTr g 4592 3972 4692 117042 0608 SSE 4309 0608 3701 6 It is customary to summarize the calculations in the following ANOVA table Source df SS MSSSdf F Treatment 1 1 2 0608 0304 99 Error 1J 1 12 3701 0308 Total 14 4309 The on 01 critical value is F0172712 693 Since the rejec tion rule speci es that H0 be rejected ifF gt F0172712 H0 is not rejected Multiple Comparisons When H0 1 2 y is rejected it is not clear which of the yi s are different from each other Methods for doing this further analysis while preserving the overall level or are called multiple comparison procedures The one we present here is recommended for deciding whether of 2 it for each i and j Tukey s procedure This depends on the socalled studentized range distribution which is characterized by a numerator df m and denominator df v Let QOWV denote the uppertail or critical value of the studentized rdistr with mv df which are given in Table A7 Tukey s procedure is based on Proposition With probability 1 0c for every i and with i 75 ij l 1 Note that the proba bility statement in the Proposition holds simultaneously for all i 7 j It follows that if for a pair 139 ji 7 j the interval MSE Xi Xj i ro1IJ1 T does not contain zero it can be concluded that M and y differ signi cantly at level OL The following steps for carrying out Tukey s procedure lead to an organized way of presenting the results from all pairwise comparisons 1 Select 06 and nd wa from Table A8 2 Calculatew QOLMU MS J 3 List the sample means in increasing order and underline each pair that differs by less than w Pairs that are not un derlined indicate that the corresponding population means differ signi cantly at level OL Example Four different concentrations of ethanol are com pared for their effect on sleep time Each concentration was given to a sample of 5 rats and the REM sleep time for each rat was recorded yieldingfl 79282 61543 47924 3276 The analysis of variance calculations are summarized in the ANOVA table Source df SS MS F Treatment 3 58823575 196078583 2109 Error 16 14874000 929625 Total 19 73697575 Since F0le 420 it follows that H0 y 4 is re jected at level or 01 To identify which of the means differ we apply Tukey s procedure 1 Q01416 519 2w519 929652238 3 54 A73 f2 f1 3276 4792 6154 7928 Thus 1 3 1 4 and 2 4 are signi cantly different at 06 01

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