Analysis of Variance Bios 662

Transcription

1 Analysis of Variance Bios 662 Michael G. Hudgens, Ph.D. mhudgens :34 BIOS ANOVA

2 Outline Introduction Alternative models SS decomposition Example w/ SAS, R BIOS ANOVA

3 Analysis of Variance Model Ch 10 text (skip ); Ch 12 How do we test hypotheses about the mean of more than 2 groups? Analysis of variance (ANOVA) model Definition 10.1 An analysis of variance model is a linear regression model in which the predictor variables are classification variables. The categories of a variable are called the levels of the variable. Categorical predictor variables are also called qualitative factors BIOS ANOVA

4 Notation Let Y ij be the j th observation in the i th group i = 1,..., K; j = 1,..., n i Let N = K i=1 n i Ȳi = j Y ij/n i BIOS ANOVA

5 ANOVA Model and Hypotheses Assume Y ij N(µ i, σ 2 ) Want to test H 0 : µ 1 = µ 2 = = µ K versus H A : at least one BIOS ANOVA

6 Two variance estimators The pooled estimate of σ 2 is: s 2 p = Ki=1 (n i 1)s 2 i Ki=1 (n i 1) Under H 0, the (weighted) variance of the Ȳi s will estimate σ 2 Ki=1 ˆσ 2 n = i (Ȳi Ȳ )2 where Ȳ = K 1 Ki=1 ni j=1 Y ij N BIOS ANOVA

7 ANOVA: F test It can be shown under H 0 : (N K)s 2 p/σ 2 χ 2 N K (K 1)ˆσ 2 /σ 2 χ 2 K 1 and s 2 p and ˆσ 2 are independent Therefore, under H 0, F ˆσ2 s 2 p F K 1,N K BIOS ANOVA

8 ANOVA: F test To test H 0, C α = {F : F > F 1 α;k 1,N K } The test uses F > F 1 α;k 1,N K because under H A, E(ˆσ 2 ) > E(s 2 p) In particular, E(s 2 p) = σ 2 whereas E(ˆσ 2 ) = σ 2 + i n i(µ i µ) 2 K 1 where µ is the overall mean defined in equation (1) below BIOS ANOVA

9 ANOVA: Example Ex: Passive smoking and lung function A study was conducted to compare the lung function of groups of smokers and non-smokers. Lung function was measured by forced mid-expiratory flow (FEF) BIOS ANOVA

10 ANOVA: Example FEF for males by smoking status Group n i Mean (L/sec) sd (L/sec) Non-smokers Passive smokers Noninhalers Light smk Mod. smk Heavy smk BIOS ANOVA

11 ANOVA: Example C.05 = {F > F 5,1044;.95 = 2.22} s 2 p = 199(.79) (.77) (.82) = ˆσ 2 = 200( ) ( ) 2 5 F = /0.636 = 58.0; Reject H 0 Reference: NEJM 302(13): 720-3, = BIOS ANOVA

12 Aside: Obtaining Quantiles/CDFs In R > qf(.95,5,1044) [1] > pf( ,5,1044) [1] 0.95 In SAS data; y = finv(.95,5,1044); y1 = quantile( F,.95,5,1044); fy = cdf( F, ,5,1044); proc print; Obs y y1 fy BIOS ANOVA

13 Cell Means Model Heretofore, we have looked at ANOVA model Y ij = µ i + ɛ ij for i = 1, 2,..., K; j = 1, 2,..., n i where ɛ ij N(0, σ 2 ) for all i, j BIOS ANOVA

14 Factor Effects Model Altenatively, an equivalent model is Y ij = µ + α i + ɛ ij for i = 1, 2,..., K; j = 1, 2,..., n i where and Ki=1 n µ = i µ i N α i = µ i µ ɛ ij N(0, σ 2 ) for all i, j (1) BIOS ANOVA

15 Factor Effects Model Note typo in text page 363 Constraint K n i α i = 0 i=1 α i does not denote type I error BIOS ANOVA

16 Model Equivalency Equivalency of null hypotheses H 0 : µ 1 = = µ K H 0 : α i = 0; i = 1, 2,..., K α i is called the i th main effect or factor effect Y ij = µ + (µ i µ) + ɛ ij = µ + α i + ɛ ij = mean + i th main effect + error Data can be partitioned similarly Y ij = Ȳ + (Ȳi Ȳ ) + (Y ij Ȳi ) = Ȳ + a i + e ij BIOS ANOVA

17 ANOVA: Sum of Squares It can be shown (below) K n i (Y ij Ȳ )2 = K n i (Ȳi Ȳ )2 + K n i (Y ij Ȳi ) 2 i=1 j=1 i=1 j=1 i=1 j=1 That is SST = SSA + SSW = (K 1)ˆσ 2 + (N K)s 2 p BIOS ANOVA

18 ANOVA: Sum of Squares Expected value of sum of squares K K E{ n i (Ȳi Ȳ )2 } = n i αi 2 + (K 1)σ2 i=1 E{ i=1 K n i (Y ij Ȳi ) 2 } = (N K)σ 2 i=1 j=1 Under H 0 : α 1 = = α K = 0, K E{ n i (Ȳi Ȳ )2 } = (K 1)σ 2 i=1 BIOS ANOVA

19 ANOVA: F test Therefore, under H A : at least one α i 0, E(F ) > 1 I.e. we reject H 0 if F is too large C α = {F : F > F 1 α;k 1,N k } BIOS ANOVA

20 ANOVA Table Source of variation df MS F Among groups K 1 ˆσ 2 = Ki=1 n i (Ȳi Ȳ )2 K 1 MSA/MSW Within groups N K s 2 p = Ki=1 (n i 1)s 2 i N K Total N 1 BIOS ANOVA

21 ANOVA: Sum of Squares Proof Start with (Y ij Y ) 2 = (Y ij Y i + Y i Y ) 2 ij ij RHS equivalent to (Y ij Y i ) 2 + (Y i Y ) 2 +2 (Y ij Y i )(Y i Y ). ij ij ij Last term can be written as 2 {(Y i Y ) (Y ij Y i )}, i j which equals zero since (Y ij Y i ) = 0 for all i. j BIOS ANOVA

22 ANOVA: E(SSW) Proof E(SSW ) = E{ ij (Y ij Y i ) 2 } = E { i (n i 1) j (Y ij Y i ) 2 n i 1 } = i (n i 1)E(s 2 i ) = i (n i 1)σ 2 = (N K)σ 2 BIOS ANOVA

23 ANOVA: Example Table 10.1: Distribution of ages (in months) at which infants first walked alone Active Passive No-Exercise Eight-week Group Group Group Control group BIOS ANOVA

24 ANOVA: Example Age Active Passive None Control Group BIOS ANOVA

25 ANOVA: SAS proc glm data=one; * proc anova data=one; class group; model age=group; run; Sum of Source DF Squares Mean Square F Value Pr > F Model Error Corrected Total BIOS ANOVA

26 ANOVA: SAS * factor effects model; data two; set one; x1=0; x2=0; x3=0; if group="active" then x1=1; if group="passive" then x2=1; if group="no" then x3=1; if group="eight" then do; x1=x2=x3=-6/5; end; proc reg data=two; model age = x1 x2 x3; Sum of Mean Source DF Squares Square F Value Pr > F Model Error Corrected Total BIOS ANOVA

27 ANOVA: R > av <- aov(age ~ group) > anova(av) Analysis of Variance Table Response: age Df Sum Sq Mean Sq F value Pr(>F) group Residuals BIOS ANOVA