Chapter 11: I = 2 samples independent samples paired samples Chapter 12: I 3 samples of equal size J one-way layout two-way layout

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1 Serk Sagtov, Chalmers and GU, February 0, 018 Chapter 1. Analyss of varance Chapter 11: I = samples ndependent samples pared samples Chapter 1: I 3 samples of equal sze one-way layout two-way layout 1 One-way layout Consder I ndependent IID samples (Y 11,..., Y 1 ),..., (Y I1,..., Y I ) measurng I treatment results. We have one man factor (factor A havng I levels) as the prncple cause of varaton n the data. The goal s to test H 0 : all I treatments have the same effect, vs H 1 : there are systematc dfferences. Data: 70 measurements of chlorphenramne maleate n tablets wth a nomnal dosage of 4 mg. Seven labs made ten measurements each: I = 7, = 10. Normal theory model Lab Mean Normally dstrbuted observatons Y j N(µ, σ ) wth equal varances (compare to the t-tests). In other words, Y j = µ + α + ɛ j, α = 0, ɛ j N(0, σ ), meanng: obs = overall mean + dfferental effect + nose. Sample means as maxmum lkelhood estmates Ȳ. = 1 Y j, Ȳ.. = 1 Y. = 1 Y j, I I j j ˆµ = Ȳ.., ˆµ = Ȳ., ˆα = Ȳ. Ȳ.., ˆα = 0, so that Y j = ˆµ + ˆα + ˆɛ j, where ˆɛ j = Y j Ȳ. are the so-called resduals Decomposton of the total sum of squares: SS T = SS A + SS E. SS T = j (Y j Ȳ..) total sum of squares for the pooled sample wth df T = I 1, SS A = ˆα factor A sum of squares (between-group varaton) wth df A = I 1, SS E = j ˆɛ j error sum of squares (wthn-group varaton) wth df E = I( 1). Mean squares and ther expected values MS A = SS A df A, E(MS A ) = σ + I 1 = SS E df E, E( ) = σ. α, 1

2 One-way F -test Pooled sample varance s p = = s an unbased estmate of σ. Use F = MS A 1 I( 1) (Y j Ȳ.) as test statstc for H 0 : α 1 =... = α I = 0 aganst H 1 : α u α v for some (u, v). Reject H 0 for large values of F, snce E H0 (MS A ) = σ and E H1 (MS A ) = σ + I 1 j α > σ. Null dstrbuton F F n1,n wth degrees of freedom n 1 = I 1 and n = I( 1). If X 1 χ n 1 and X χ n are two ndependent random varables, then X 1/n 1 X /n F n1,n. The normal probablty plot of resduals ˆɛ j supports the normalty assumpton. Nose sze σ s estmated by s p = = One-way Anova table Source df SS MS F P -value Labs Error Total Whch of the ( 7 ) = 1 parwse dfferences are sgnfcant? Usng the 95% CI for a sngle par of ndependent samples (µ u µ v ) we get (Ȳu. Ȳv.) ± t 63 (0.05) sp = (Ȳu. Ȳv.) ± 0.055, 5 where t 63 (0.05) =.00. Ths formula yelds 9 sgnfcant dfferences: Labs Dff The multple comparson problem: the above CI formula s amed at a sngle dfference, and may produce false dscoveres. We need a smultaneous CI formula for all 1 parwse comparsons. Bonferron method Thnk of k ndependent replcatons of a statstcal test. The overall result s postve f we get at least one postve result among these k tests. The overall sgnfcance level α s obtaned, f each sngle test s performed at sgnfcance level α/k: ndeed, assumng the null hypothess s true, the number of postve results s X Bn(k, α k ), and due to ndependence P(X 1 H 0 ) = 1 (1 α k )k α for small values of α. Smultaneuos 100(1 α)% CI formula for ( I ) parwse dfferences (µu µ v ):

3 (Ȳu. Ȳv.) α ± t I( 1) ( I(I 1) ) s p Flexblty of the formula: works for dfferent sample szes as well after replacng by Warnngs: ( I ) parwse Anova comparsons are not ndependent as requred by Bonferron method, Bonferron method gves narrower ntervals compared to the Tukey method. 1 u + 1 v. The Bonferron smultaneuos 95% CI for (α u α v ) (Ȳu. Ȳv.) ± t 63 (.05) sp 4 5 = (Ȳu. Ȳv.) ± 0.086, where t 63 (0.001) = 3.17, detects 3 sgnfcant dfferences between labs (1,4), (1,5), (1,6). Tukey method If I ndependent samples (Y 1,..., Y ) taken from N(µ, σ ) have the same sze, then the sample means Ȳ. N(µ, σ ) are ndependent. Consder the range of dfferences between (Ȳ. µ ): Then we get R(I; ) = max{ȳ1. µ 1,..., ȲI. µ I } mn{ȳ1. µ 1,..., ȲI. µ I }. R(I; ) s p / SR(I, I( 1)), where the so-called studentzed range dstrbuton SR(k, df) has two parameters: the number of samples k, and the number of degrees of freedom used n the varance estmate s p. Tukey s 95% smultaneuos CI = (Ȳu. Ȳv.) ± q I,I( 1) (0.05) sp Usng q 7,60 (0.05) = 4.31 from the SR-dstrbuton table, we fnd four sgnfcant parwse dfferences: (1,4), (1,5), (1,6), (3,4), snce (Ȳu. Ȳv.) ± q 7,63 (0.05) = (Ȳu. Ȳv.) ± Kruskal-Walls test A nonparametrc test, wthout assumng normalty, for H 0 : all observatons are equal n dstrbuton, no treatment effects. Extendng the dea of the rank-sum test, consder the pooled sample of sze N = I. Let R j be the pooled ranks of the sample values Y j, so that j R j = N(N+1) and R.. = N+1 s the mean rank. Kruskal-Walls test statstc K = 1 I N(N+1) =1 ( R. N+1 ) 3

4 Reject H 0 for large K usng the null dstrbuton table. For I = 3, 5 or I 4, 4, use the approxmate null dstrbuton K a χ I 1. In the table below the actual measurements are replaced by ther ranks Wth the observed test statstc K = 8.17 and df = 6, usng χ 5-dstrbuton table we get a P-value Two-way layout Labs Means Suppose the data values are nfluenced by two man factors and a nose: Y jk = µ + α + β j + δ j + ɛ jk, = 1,..., I, j = 1,...,, k = 1,..., K, grand mean + man A-effect +man B-effect + nteracton + nose. Factor A has I levels, factor B has levels, and we have K observatons for each combnaton (, j). Normal theory model Key assumpton: all nose components ɛ jk N(0, σ ) are ndependent and have the same varance. Parameter constrants and numbers of degrees of freedom df A = I 1, because α = 0, df B = 1, because j β j = 0, df AB = I I + 1 = (I 1)( 1), because δ j = 0, j δ j = 0. Maxmum lkelhood estmates: ˆµ = Ȳ..., ˆα = Ȳ.. Ȳ..., ˆβj = Ȳ.j. Ȳ..., ˆδ j = Ȳj. Ȳ... ˆα ˆβ j = Ȳj. Ȳ.. Ȳ.j. + Ȳ..., and the resduals ˆɛ jk = Y j. Ȳjk. Example (ron retenton) Raw data X jk s the percentage of ron retaned n mce. Factor A: I = ron forms, factor B: = 3 dosage levels, K = 18 observatons for each (ron form, dosage level) combnaton. From the graphs we see that the raw data s not normally dstrbuted. However, the transformed data Y jk = ln(x jk ) produce more satsfactory graphs. The sample means and maxmum lkelhood estmates for the transformed data 4

5 ( ) (Ȳj.) = two rows produce two profles: not parallel - possble nteracton, Ȳ... = 1.9, ˆα 1 = 0.14, ˆα = 0.14, ( ) ˆβ 1 = 0.50, ˆβ = 0.08, ˆβ3 = 0.4, (ˆδ j ) = Sums of squares SS T = j k (Y jk Ȳ...) = SS A + SS B + SS AB + SS E, df T = IK 1 SS A = K ˆα, df A = I 1, MS A = SS A df A, E(MS A ) = σ + K I 1 α j β j SS B = IK ˆβ j, df B = 1, MS B = SS B df B, E(MS B ) = σ + IK 1 SS AB = K j δ j, df AB = (I 1)( 1), MS AB = SS AB df AB, E(MS AB ) = σ + SS E = j k (Y jk Ȳj.), df E = I(K 1), = SS E df E, E( ) = σ Three F -tests Pooled sample varance s p = s an unbased estmate of σ. K (I 1)( 1) j δ j Null hypothess No-effect property Test statstcs and null dstrbuton H A : α 1 =... = α I = 0 E(MS A ) = σ F A = MS A F dfa,df E H B : β 1 =... = β = 0 E(MS B ) = σ F B = MS B F dfb,df E H AB : all δ j = 0 E(MS AB ) = σ F AB = MS AB F dfab,df E Reject null hypothess for large values of the respectve test statstc F. Inspect normal probablty plot for the resduals ˆɛ jk. Example (ron retenton) Two-way Anova table for the transformed ron retenton data. Dosage effect was expected from the begnnng. Interacton s not sgnfcant. Source df SS MS F P Iron form Dosage Interacton Error Total Sgnfcant effect due to ron form. Estmated log scale dfference ˆα ˆα 1 = Ȳ.. Ȳ1.. = 0.8 yelds the multplcatve effect of e 0.8 = 1.3 on a lnear scale. 3 Randomzed block desgn Blockng s used to remove the effects of a few of the most mportant nusance varables. Randomzaton s then used to reduce the contamnatng effects of the remanng nusance varables. Block what you can, randomze what you cannot. 5

6 Expermental desgn: randomly assgn I treatments wthn each of blocks. Test the null hypothess of no treatment effects usng the two-way layout Anova. The block effect s antcpated and s not of major nterest. Examples: Block Treatments Observaton A homogeneous plot of land I fertlzers each appled to The yeld on the dvded nto I subplots a randomly chosen subplot subplot (, j) A four-wheel car 4 types of tres tested on the same car tre s lfe-length A ltter of I anmals I dets randomly assgned to I snlngs the weght gan Addtve model If K = 1, then we cannot estmate nteracton. Ths leads to the addtve model wthout nteracton Y j = µ + α + β j + ɛ j. Maxmum lkelhood estmates ˆµ = Ȳ.., ˆα = Ȳ. Ȳ.., ˆβ = Ȳ.j Ȳ.., ˆɛ j = Y j Ȳ.. ˆα ˆβ = Y j Ȳ. Ȳ.j + Ȳ.. Sums of squares SS T = j (Ȳj Ȳ..) = SS A + SS B + SS E, df T = I 1 SS A = ˆα, df A = I 1, MS A = SS A df A SS B = I ˆβ j j, df B = 1 MS B = SS B SS E = df B F A = MS A F B = MS B j ˆɛ j, df E = (I 1)( 1) = SS E df E E( ) = σ Example (tchng) Data: the duraton of the tchng n seconds Y j, wth K = 1 observaton per cell, I = 7 treatments to releve tchng appled to = 10 male volunteers aged F dfa,df E F dfb,df E Subject No Drug Placebo Papaverne Morphne Amnophyllne Pentabarbtal Trpelennamne BG F BS SI BW TS GM SS MU OS Boxplots ndcate volatons of the assumptons of normalty and equal varance. Notce much bgger varance for the placebo group. Two-way Anova table Source df SS MS F P Drugs Subjects Error Total

7 Tukey s method of multple comparson q I,(I 1)( 1) (α) sp 3096 = q 7,54 (0.05) = 75.8 reveals only 10 one sgnfcant dfference: papaverne vs placebo wth = 90. > Fredman test Treatment Mean Nonparametrc test, when ɛ j are non-normal, to test H 0 : no treatment effects. Rankng wthn j-th block: (R 1j,..., R Ij ) = ranks of (Y 1j,..., Y Ij ) so that R 1j R Ij = I(I+1) mplyng 1(R I 1j R Ij ) = I+1 and R.. = I+1. Test statstc Q = 1 I I(I+1) =1 ( R. I+1 ) has an approxmate null dstrbuton Q a χ I 1. Snce Q s a measure of agreement between rankngs, we reject H 0 for large values of Q. Example (tchng) From the values R j and R. below and I+1 = 4, we fnd the Fredman test statstc Q = Usng the ch-square dstrbuton table wth df = 6 we obtan an approxmate P-value to be.14%. We reject the null hypothess of no effect even n the non-parametrc settng. Subject No Drug Placebo Papaverne Morphne Amnophyllne Pentabarbtal Trpelennamne BG F BS SI BW TS GM SS MU OS R , 7