Topic 3. Single factor ANOVA: Introduction [ST&D Chapter 7] 3.1. The F distribution [ST&D p. 99]

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1 Topic 3. Single faco ANOVA: Inoducion [ST&D Chape 7] "The analyi of vaiance i moe han a echnique fo aiical analyi. Once i i undeood, ANOVA i a ool ha can povide an inigh ino he naue of vaiaion of naual even" Sokal & Rohlf (1995, BIOMETRY The F diibuion [ST&D p. 99] Aume ha you ae ampling a andom fom a nomally diibued populaion (o fom wo diffeen populaion wih equal vaiance by fi ampling n 1 iem and calculaing hei vaiance 1 (df: n 1-1, followed by ampling n iem and calculaing hei vaiance (df: n - 1. Now conide he aio of hee wo ample vaiance: 1 Thi aio will be cloe o 1, becaue hee vaiance ae eimae of he ame quaniy. The expeced diibuion of hi aiic i called he F-diibuion. The F-diibuion i deemined by wo value fo degee of feedom, one fo each ample vaiance. Saiical Table fo F (e.g. A6 in you book how he cumulaive pobabiliy diibuion of F fo eveal eleced pobabiliy value. The value in he able epeen F [1, ] whee i he popoion of he F-diibuion o he igh of he given- F-value (in one ail and 1, ae he degee of feedom peaining o he numeao and denominao of he vaiance aio, epecively F(1,40 F(8,6 F(6, F Figue 1 Thee epeenaive F-diibuion (noe imilaiy of F (1,40 o 1. Fo example, a value F /=0.05, 1=9, = 9] = 4.03 indicae ha he aio 1 /, fom ample of en individual fom nomally diibued populaion wih equal vaiance, i expeced o be 1

2 lage han 4.03 by chance in only 5% of he expeimen (he alenaive hypohei i 1 o i i a wo ail e. 3.. Teing he hypohei of equaliy of vaiance [ST&D ] Suppoe X 1,..., X m ae obevaion dawn fom a nomal diibuion wih mean μ X and vaiance σ x ; and Y 1,..., Y n ae dawn fom a nomal diibuion wih mean μ v and vaiance σ y. The F aiic can be ued a a e fo he hypohei H 0 : σ x = σ y v. he hypohei H 1 : σ x σ y. H 0 i ejeced a he level of ignificance if he aio x = y i eihe F /, dfx-1, dfy-1 o F 1-/, dfx-1, dfy-1. In pacice, hi e i aely ued becaue i i vey eniive o depaue fom nomaliy. Thi can be calculaed uing SAS PROC TTEST Teing he hypohei of equaliy of wo mean [ST&D 98-11] The aio beween wo eimae of can be ued o e diffeence beween mean, ha i, a e of H 0 : 1 - = 0 veu H 1 : 1-0. In paicula: F = eimae of σ fom ample mean eimae of σ fom individual The denominao i an eimae of fom he individual wihin each ample. Tha i, i i a weighed aveage of he ample vaiance. The numeao i an eimae of povided by he vaiaion among ample mean. To obain hi eimae of (vaiance among individual fom he mean vaiance ( Y n, o be muliplied by n. need Y F among wihin n Y When he wo populaion have diffeen mean (bu ame vaiance, he eimae of baed on ample mean will include a conibuion aibuable o he diffeence beween populaion mean a well a any andom diffeence (i.e. wihin-populaion vaiance. Thu, if hee i a ignifican diffeence among mean, he ample mean ae expeced o be moe vaiable han when chance alone opeae and hee ae no ignifican diffeence among mean.

3 Nomenclaue: o be conien wih he book in he ANOVA we will ue fo he numbe of eplicaion and n fo he oal numbe of expeimenal uni in he expeimen Example: We will explain he e uing a daa e of Lile and Hill (p. 31. Table 1. Yield (100 lb./ace of whea vaieie 1 and fom plo o which he vaieie wee andomly aigned: Vaieie Replicaion Y i. Y i. i Y 1. = Y. = Y.. = 185 Y.. = 18.5 In hi expeimen, hee ae wo eamen level ( = and five eplicaion ( = 5. Each obevaion in he expeimen ha a unique "adde" given by Y ij, whee i i he index fo eamen (i = 1 o and j i he index fo eplicaion (j = 1,, 3, 4, o 5. Thu Y,4 = 19. The do noaion i a hohand alenaive o uing. Summaion i fo all value of he ubcip occupied by he do. Thu Y 1. = and Y. = Fo hi execie, we will aume ha he wo populaion have he ame (unknown vaiance σ and hen e H 0 : μ 1 = μ. We do hi by obaining wo eimae fo σ and compaing hem. Fi, we can compue he aveage ample vaiance σ wihin ample. To deemine hi vaiabiliy called expeimenal eo, we compue he vaiance of each ample ( 1 and, aume hey boh eimae a common vaiance, and hen eimae ha common vaiance by pooling he amplevaiance: 1 j ( Y 1 j 1 Y 1 1., j ( Y j Y 1. ( ( 1 pooled = 4* * 4.0 / (4 + 4 = 5.5 ( 1 ( 1 1 wihin In hi cae, ince n 1 = n, he pooled vaiance i imply he aveage of he wo ample vaiance. Since pooling 1 and give an eimae of σ baed on he vaiabiliy wihin ample, we will deignae i w (ubcip w= wihin. The econd eimae of he ample vaiance σ i baed on he vaiaion beween ample ( b. Auming ha hee wo ample ae andom ample dawn fom he ame populaion and ha, heefoe, Y 1. and Y. boh eimae he ame populaion mean, we eimae he vaiance of mean uing Y. Recall fom Topic 1 ha he mean Y of a e of andom vaiable dawn fom 3

4 a nomal diibuion wih mean μ and vaiance σ i ielf a nomally diibued andom vaiable wih mean μ and vaiance σ /. The fomula fo Y i Y ( Yi. Y.. i1 1 = [( ( ] / (-1 = 4.5 and, fom he cenal limi heoem, n ime Y povide an eimae fo σ (n i he numbe of vaiae on which each ample mean i baed. Theefoe, he beween ample eimae i: Y = 5 * 4.5 =.5 b Thee wo vaiance ae ued in he F e a follow. If he null hypohei i no ue, hen he beween ample vaiance hould be much lage han he wihin ample vaiance ("much lage" mean lage han one would expec by chance alone. Theefoe, we look a he aio of hee vaiance and ak whehe hi aio i ignificanly geae han 1. I un ou ha unde ou aumpion (nomaliy, equal vaiance, ec., hi aio i diibued accoding o an F (-1, (- 1 diibuion. Tha i, we define: F = b / w and e whehe hi aiic i ignificanly geae han 1. The F aiic meaue how many ime lage i he vaiabiliy beween he ample compaed wih he vaiabiliy wihin ample. In hi example, F =.5/5.5 = 4.9. The numeao b i baed on 1 df, ince hee ae wo ample mean. The denominao, w, i baed on pooling he df wihin each ample o df w = (-1 = (4 = 8. Fo hee df, we would expec an F value of 4.9 o lage ju by chance abou 7% of he ime. Fom Table A.6 (p.614 of ST&D, F 0.05, 1, 8 = 5.3. Since 4.9 < 5.3, we fail o ejec H 0 a he 0.05 ignificance level Relaionhip beween F and In he cae of only wo eamen, he quae-oo of he F aiic i diibued accoding o a diibuion: F 1, df 1, ( 1 meaning 1, df ( 1 b w 4

5 In he example above, wih 5 ep pe eamen: F (1,8, 1 - = ( 5, 1- / (be caeful: F ue and / The oal degee of feedom fo he aiic i ( - 1 = - = n- ince hee ae n obevaion and hey mu aify conain equaion, one fo each eamen mean. Theefoe, we ejec he null hypohei a he ignificance level if > /, (-1. Hee ae he compuaion fo ou daa e: b w Since.07 < 0.05, 8 =.306, we fail o ejec H 0 a he 0.05 ignificance level. The value.306 i obained fom Table A3 (p.611 and df= (5-1= 8. Noe ha.306 = 5.3= F 0.05, 1, The linea addiive model [ST&D p. 3, 103, 15] One populaion: In aiic, a common model decibing he makeup of an obevaion ae ha i coni of a mean plu an eo. Thi i a linea addiive model. A minimum aumpion i ha he eo ae andom, making he model pobabiliic ahe han deeminiic. The imple linea addiive model: Y i = + i Thi model i applicable o he poblem of eimaing o making infeence abou populaion mean and vaiance. Thi model aemp o explain an obevaion Y i a a mean plu a andom elemen of vaiaion i. The i ' ae aumed o be fom a populaion of uncoelaed ' wih mean zeo. Independence among ' i aued by andom ampling Two populaion: Thi econd model i moe geneal han he peviou model (3.4.1 becaue i pemi u o decibe wo populaion imulaneouly: Y ij = + i + ij Fo ample fom wo populaion wih poibly diffeen mean bu a common vaiance, any given obevaion i compoed of: 5

6 he gand mean of he populaion, a componen i fo he populaion involved (i.e. + 1 = 1 and + =, and a andom deviaion ij. A befoe, he ' ae aumed o be fom a ingle populaion wih nomal diibuion, mean = 0, and vaiance σ. The ubindex i (= 1, indicae he eamen numbe and he ubindex j (= 1,..., n indicae he numbe of obevaion fom each populaion (eplicaion. The I ae alo efeed a he eamen effec, meaued a a deviaion fom each eamen mean fom an oveall mean fo he complee expeimen. Thi oveall mean i e a a middle efeence poin a: = ( 1 + /, which i eimaed by Y.. = (Y 1 + Y / Theefoe 1 + = 0 o in a diffeen way - 1 = (he diffeence beween mean i *. If 1 we e = 0. Anohe way o expe hi model, uing he do noaion fom befoe, i: Yij =Y.. + (Y i. -Y.. + (Yij - Y i Moe han wo populaion. One-way claificaion ANOVA A wih he ample -e, he linea model i: Y ij = + i + ij whee now i = 1,..., and j = 1,...,. Again, he ij ae aumed o be dawn fom a nomal diibuion wih mean 0 and vaiance. Two diffeen kind of aumpion can be made abou he ' ha will diffeeniae he Model I ANOVA fom he Model II ANOVA. The Model I ANOVA o fixed model: In hi model, he ' ae fixed and i = 0 Seing he i = 0 i imply meauing eamen effec a deviaion fom an oveall mean fom he expeimen. The null hypohei i hen H 0 : 1 =... = i = 0 6

7 and he alenaive a H 1 : a lea one i 0. Wha a Model I anova e i he diffeenial effec of eamen ha ae fixed and deemined by he expeimene. The wod "fixed" efe o he fac ha each eamen i aumed o alway have he ame effec i. The ' ae aumed o coniue a finie populaion and ae he paamee of inee, along wih. When he null hypohei i fale (and ome i 0, hee will be an addiional componen of vaiaion due o eamen effec equal o: i 1 Since he i ae meaued a deviaion fom a mean, hi quaniy i analogou o a vaiance bu canno be called uch ince i i no baed on a andom vaiable bu ahe on delibeaely choen eamen. The Model II ANOVA o andom model: In hi model, he added effec fo each goup (' ae no fixed eamen bu ae andom effec. In hi cae, we have no delibeaely planned o fixed he eamen fo any goup, and he effec on each goup ae andom and only paly unde ou conol. The ' ae a andom ample fom a populaion of ' fo which he mean i zeo and he vaiance i. When he null hypohei i fale, hee will be an addiional componen of vaiance equal o. Since he effec ae andom, i i fuile o eimae he magniude of hee andom effec fo any one goup, o he diffeence fom goup o goup. Howeve, we can eimae hei vaiance, he added vaiance componen among goup:. We e fo i peence and eimae i magniude, a well a i pecenage conibuion o he vaiaion (calculaed in SAS wih PROC VARCOMP. The null hypohei in he andom model i aed a H 0 : = 0 veu H 1 : 0. An impoan poin i ha he baic eup of daa, a well a he compuaion and ignificance e, in mo cae i he ame fo boh model. The pupoe diffe beween he wo model, a do ome of he upplemenay e and compuaion following he iniial ignificance e. In he fixed model, we daw infeence abou paicula eamen; in he andom model, we daw an infeence abou he populaion of eamen. Unil Topic 10, we will deal only wih he fixed model. Aumpion of he model [ST&D p.174] 1. Teamen and envionmenal effec ae addiive. Expeimenal eo ae andom, independenly and nomally diibued abou zeo mean and wih a common vaiance. Effec ae addiive Thi mean ha all effec in he model (eamen effec, andom eo caue deviaion fom he oveall mean in an addiive manne (ahe han, fo example, muliplicaive. 7

8 Eo em ae independenly and nomally diibued Thi mean hee i no coelaion beween expeimenal gouping of obevaion (e.g. by eamen level and he ize of he eo em. Thi could be violaed if, fo example, eamen ae no aigned andomly. Vaiance ae homogeneou Thi aumpion mean ha he vaiance of he diffeen eamen goup ae he ame. Thi aumpion mean ha he mean and vaiance of eamen hae no coelaion, ha i, ha eamen wih lage mean do no have lage vaiance. We need hi aumpion ince we ae calculaing an oveall ample vaiance, by aveaging he vaiance of he diffeen eamen. Thee ae alenaive aiical analye when he vaiance ae no homogeneou (e.g. Welch' vaiance-weighed one-way ANOVA 3.5. ANOVA: Single faco deign The Compleely Random Deign CRD In ingle faco expeimen, a ingle faco i vaied o fom he diffeen eamen. The expeimen hown below i aken fom page 141 of ST&D. The expeimen involve inoculaing five diffeen culue of one legume, clove, wih ain of he niogen-fixing baceia fom anohe legume, alfalfa. A a o of conol, a ixh ial wa un in which a compoie of he five clove culue wa inoculaed. Thee ae 6 eamen ( = 6 and each eamen i given 5 eplicaion ( = 5. Table 1. Inoculaion of clove wih Rhizobium ain [ST&D Table 7.1] 3DOK1 3DOK5 3DOK4 3DOK7 3DOK13 compoie Toal Y ij = Y i = Y.. Y ij Y i. / (Y ij - Y i Y i. = mean n-1 vaiance

9 The compleely andomized deign (CRD i he baic ANOVA deign. I i ued when hee ae diffeen eamen level of a ingle faco (in hi cae, Rhizobium ain. Thee eamen ae applied o independen andom ample of ize n. Teamen can have diffeen numbe of eplicaion, bu he fomula become much moe complicaed and we will popone hem ill lae in he coue. Le he oal ample ize fo he expeimen be deignaed a n =. Le Y ij denoe he j h meauemen (eplicaion ecoded fom he i h eamen. WARNING: ome ex inechange he i and he j (i.e. he ow and column of he able. We wih o e he hypohei H 0: 1 = = 3 =... = again H 1 : no all he i ' ae equal. Thi i a aighfowad exenion of he wo-ample e of opic 3.3 ince hee wa nohing pecial abou he value =. Recall ha he e aiic wa: In ou do noaion, we can wie hi a: F = b / w ( Yij Y i. i1 j1 SSE w, whee SSE ( 1 ( 1 i1 j1 ( Y ij Y i. Hee SSE i he um of quae fo eo. Alo: b i1 ( Y i. Y 1.. SST, whee SST= ( Y i Y 1 i1... Hee SST i he um of quae fo eamen (SAS efe o hi a he Model SS. Since he vaiance among eamen mean eimae /, he muliplicaion fo in he definiion fomula fo SST i equied o ha he mean quae fo eamen (MST will be an eimae of ahe han /. In he example 3.3 above we alo muliplied by in ode o eimae beween ample vaiance ( b = Y. In hi noaion we can wie (emembe =n: F SST /( 1 SSE /( ( 1 SST /( 1 SSE /( n We can hen define: 9

10 The mean quae fo eo: MSE = SSE/(n- give he aveage dipeion of he iem aound hei epecive goup mean. The df i ime (-1, which i he df wihin each of he pooled eamen. MSE i an eimae of a common, he expeimenal eo (= wihin vaiaion o vaiaion among obevaion eaed alike. MSE i a valid eimae of he common if he aumpion of equal vaiance among eamen i ue (becaue we ae aveaging he vaiance eimae fom diffeen eamen. The mean quae fo eamen: MST = SST/(-1. (MS Model in SAS Thi i an independen eimae of, when he null hypohei i ue (H 0 : μ 1 = μ = μ 3 =... = μ. If hee ae diffeence among eamen mean, hee will be an added componen of vaiaion due o eamen effec equal o i /(-1 (Model I o (Model II (ee opic and ST&D 155. The muliplicaion by i equied o expe he vaiance pe individual, no pe mean (emembe ha o Y Y. F = MST/MSE The F value i obained by dividing he eamen mean quae by he eo mean quae. We expec o find F appoximaely equal o 1. In fac, howeve, he expeced aio i: MST MSE i /( I i clea fom hi fomula, ha he F-e i eniive o he peence of he added componen of vaiaion due o eamen effec. In ohe wod, he ANOVA pemi u o e whehe hee ae any added eamen effec. Tha i, o e whehe a goup of mean can be conideed andom ample fom he ame populaion o whehe we have ufficien evidence o conclude ha he eamen ha have affeced each goup epaaely have euled in hifing hee mean ufficienly o ha hey can no longe be conideed ample fom he ame populaion. 1 Recall ha he numbe of degee of feedom i he numbe of independen quaniie in he aiic. Thu SST ha he quaniie (Y i. - Y.. which have one conain (ha hey mu um o 0; o df = -1. The SSE ae n quaniie Yij, which have conain fo he ample mean; o df e = (-1 = n-. We can alo ue he following equaion: 10

11 i1 j1 ( Y ij Y.. i1 ( Y i. Y.. i1 j1 ( Y ij Y i. o TSS = SST + SSE whee TSS i he oal um of quae of he expeimen In ohe wod, um of quae ae pefecly addiive. If you expand he quaniy on he lef-hand ide of he above equaion in ou do noaion, hee i a co poduc em of he fom (Y ij Y.. ha hould appea. I un ou, ha all of hee co poduc em cancel each ohe ou. Quaniie ha aify hi ae aid o be ohogonal. Anohe way of aying hi i ha we can decompoe he oal SS ino a poion due o vaiaion among goup and anohe independen poion due o vaiaion wihin goup. The degee of feedom ae alo addiive (i.e. df To = df T + df e. The do noaion above povide he "definiion" fomula fo of hee quaniie (TSS, SST, and SSE. Bu each alo ha a fiendlie "calculaion" fom o compue hem by hand. The acual calculaion, when done by hand, ue he fomula C TSS SST i1 i1 j1 Y i. Y ij ( Y.. / n ( Yij / n The coecion em (C. I he quaed um of all obevaion divided by hei numbe. ij C / C The oal um of quae ha include all ouce of vaiaion. Thi i he oal SS. The um of quae aibuable o he vaiable of claificaion. Thi i he beween SS, o among goup SS o eamen SS. SSE TSS SST The um of quae among individual eaed alike. Thi i he wihin goup SS, o eidual SS o eo SS. I i eaie o calculae a a diffeence An ANOVA able povide a yemaic peenaion of eveyhing we've coveed unil now. The fi column of he ANOVA able pecifie he componen of he linea model. The nex column indicae he df aociaed wih each of hee componen. Nex i a column wih he SS aociaed wih each, followed by a column wih he coeponding mean quae. Mean quae ae eenially vaiance; and hey ae found by dividing SS by hei epecive df. Finally, he la column in an ANOVA able below peen he F aiic, which i a aio of mean quae (i.e. a aio of vaiance. Uually, a la column i added indicaing he pobabiliy of finding ha F value by chance. An ANOVA able (including an addiional column of he SS definiional fom: 11

12 Souce df Definiion SS MS F Teamen - 1 SST SST/(-1 MST/MSE ( Y i. Y.. i Eo Toal n - 1 (-1 = n - i, j i, j ( Y ij Y i ( Y ij Y... TSS - SST TSS SSE/(n- The ANOVA able fo ou Rhizobium expeimen would look like hi: Souce df SS MS F Among culue ** Wihin culue Toal Noice ha he MSE (11.79 i he pooled vaiance o he aveage of vaiance wihin each eamen (i.e. MSE = i / ; whee i i he vaiance eimaed fom he ih eamen. The F value of 14 indicae ha he vaiaion among eamen i ove 14 ime lage han he aveage vaiaion wihin eamen. Thi value fa exceed he ciical F value fo uch an expeimenal deign a α = 0.05 (F ci = F (5,4,0.05 =.6, o we ejec H 0 and conclude ha a lea one of he eamen ha a nonzeo effec on he epone vaiable, a he pecified ignificance level Aumpion aociaed wih ANOVA The aumpion aociaed wih ANOVA can be expeed in em of he following aiical model: Y ij = + i + ij. Fi, ij (he eidual ae aumed o be nomally diibued wih mean 0 and poe a common vaiance σ, independen of eamen level i and ample numbe j. Noe ha he ANOVA equie ha he eidual eo have a nomal diibuion and no ha he complee populaion of oiginal value have a nomal diibuion. If hee i a eamen effec, mixing he diffeen eamen will eul in a diibuion wih muliple peak. Theefoe, he eamen effec ae fi ubaced, he eidual ij ae calculaed and hen hei nomaliy i eed. 1

13 Nomal diibuion Recall fom he fi lecue ha he Shapio-Wilk e aiic W (ST&D 567; poduced by SAS via Poc UNIVARIATE NORMAL povide a poweful e fo nomaliy fo mall o medium ample (n < 000. Nomaliy i ejeced if W i ufficienly malle han 1. W i imila o a coelaion beween he daa and hei nomal coe (ST&D 566. In a pefecly nomal populaion hee i a pefec coelaion W=1. Fo lage populaion (n>000, SAS ecommend he ue of he Kolmogoov-Sminov aiic (ST&D 571; alo poduced by SAS via Poc UNIVARIATE NORMAL o via Analy. Boh e ae applied o he eidual of he model, which ae eay o calculae in SAS o R Homogeneiy of vaiance Te fo homogeneiy of vaiance (i.e. homocedaiciy aemp o deemine if he vaiance i he ame wihin each of he goup defined by he independen vaiable. Bale' e (ST&D 481 can be vey inaccuae if he undelying diibuion i even lighly nonnomal, and i i no ecommended fo ouine ue. Levene' e i moe obu o deviaion fom nomaliy, and will be ued in hi cla. Levene e i an ANOVA of he quae of he eidual of each obevaion fom i eamen mean. An alenaive fom of he e, implemened in R, ue he abolue value of he deviaion fom he eamen median. To pefom Levene e in SAS, you need o ue he opion HOVTEST (fo Homogeneiy of vaiance e wihin he mean aemen in he PROC GLM pocedue: poc GLM; Cla Teamen; Model Repone = Teamen; Mean Teamen / Hove = Levene; If Levene' e ejec he hypohei of homogeneiy of vaiance hee ae hee alenaive: 1. Tanfom he daa (e.g. logaihm o ha he anfomed value have unifom vaiance.. Ue a non paameic aiical e. 3. Ue he WELCH opion which poduce a Welch' vaiance-weighed ANOVA (Biomeika 1951 v38, 330 inead of he uual ANOVA. Thi alenaive o he uual analyi of vaiance i moe obu if vaiance ae no equal. poc GLM; Cla Teamen; Model Repone = Teamen; Mean Teamen / Welch; 13

14 Expeimenal Pocedue: Randomizaion Hee i how he clove plo migh look if hi expeimen wee conduced in he field: The expeimenal pocedue would be: Fi, andomly (e.g. fom a andom numbe able uch a ST&D 606, o uing PROC PLAN in SAS, ec. elec he plo numbe o be aigned o he ix eamen (A, B, C, D, E, F. Example: On p. 607, aing fom Row 0, column (a andom aing poin, move downwad. Take fo eamen A he fi 5 andom numbe unde 30, and o foh (wihou eplicaion: Teamen A: 5, 19, 13, 0, 6; Teamen B: 14, 6, 1, 8, 4; ec. O imply wie 30 numbe, mixed and andomly aigned 5 numbe o each eamen B 3 B A A 7 B A B A A B Powe and ample ize Peaon and Haley (1953, Biomeika 38: povided powe funcion cha ha ae eay o ue o calculae he powe of an ANOVA and he appopiae numbe of eplicaion. The Table ae available in he cla webie a hp:// Thee ae diffeen cha fo each diffeen numeao degee of feedom Powe The powe of a e i he pobabiliy of deecing a nonzeo eamen effec. To calculae he powe of he F e in an ANOVA uing Peaon and Haley' powe funcion cha, i i neceay o calculae fi a ciical value. Thi ciical value depend on he numbe of eamen (, he numbe of eplicaion (n, he magniude of he eamen effec ha he 14

15 inveigao wihe o deec (d, an eimae of he populaion vaiance ( = MSE, and he pobabiliy of ejecing a ue null hypohei (. In a CRD, y ij = µ + i + ij, i = 1,,,; j = 1,,...,; µ i he oveall mean; i i he eamen effec ( i =µ i - µ. To calculae he powe, you fi need o calculae, a andadized meaue (in uni of he expeced diffeence among mean which can be ued o deemine ample ize fom he powe cha. The exac fomula i: MSE i Thi geneal fomula can be implified uing an appoximaion ha aume all i ae zeo excep he wo exeme eamen effec (le' call hem K and L, o ha d = µ K - µ L. You can hink of d a he diffeence beween he exeme eamen mean. Taking µ o be in he middle of µ K and µ L, i = d/: i ( d / ( d / d / 4 d / 4 d / d And he appoximae fomula implifie: d * * MSE Fi elecing he cha fo 1 = df 1 = df numeao = df eamen = -1 and hen chooe he x-axi cale fo he appopiae (0.05 o Daw an imaginay veical line a he calculaed and look fo he inecepion wih he cuve fo = df = df denominao = df eo = (-1. The coeponding value a he y-axi give he powe of he e. Example: Suppoe ha one expeimen had = 6 eamen wih = eplicaion each. The diffeence beween he exeme mean wa 10 uni, MSE= 5.46, and he equied = 5%. To calculae he powe uing he appoximae fomula: d * * MSE 10 * (6* Ue Cha 1 = -1 = 5 and he e of cuve o he lef ( = 5%. Selec cuve = (n-1 = 6. The heigh of hi cuve coeponding o he abcia of = 1.75 i he powe of he e. In hi 15

16 cae, he powe i lighly geae han Expeimen hould be deigned o have a powe of a lea 80% (i.e. β 0.0. To calculae he powe uing Analy: Saiic ANOVA One-Way ANOVA Te Powe analyi. O Analy Sample Size One-Way ANOVA Complee he numbe of eamen, he coeced um of quae CSS (= SST = beween SS = among goup SS = eamen SS, and he andad deviaion, which i he quae oo of he mean quaed eo (MSE. You mu alo pecify he ignificance level of he e; he defaul i Sample ize To calculae he numbe of eplicaion fo a given and deied powe: Specify he conan. Sa wih an abiay o compue. Ue he appopiae Peaon and Haley cha o find he powe. Ieae he poce unil a minimum value which aifie he equied powe fo a given level i found. Example: Suppoe ha 6 eamen will be involved in a udy and he anicipaed diffeence beween he exeme mean i 15 uni. Wha i he equied ample ize o ha hi diffeence will be deeced a = 1% and powe = 90%, knowing ha = 1? (noe, = 6, = 1%, = 10%, d = 15, and MSE = = 1. df (1- fo =1% 6(-1= (3-1= (4-1= Thu 4 eplicaion ae equied fo each eamen o aify he equied condiion Subampling: he need deign [ST&D p ] I may happen ha he expeimene wihe o make eveal obevaion wihin each expeimenal uni, he uni o which he eamen i applied. Such obevaion ae made on ubample o ampling uni. The claical example of hi i given in Seel and Toie: ampling individual plan wihin po whee he po ae he expeimenal uni andomly aigned o eamen. Ohe example would be individual ee wihin an ochad plo (whee he eamen i aigned o he plo, individual heep wihin a hed (whee he eamen i aigned o he hed, ec. We call he analyi of hi kind of daa oganized in a hieachical way need analyi of vaiance. Need ANOVA ae no limied o wo hieachical level (e.g. po, and hen plan wihin po. We can divide he ubgoup ino ub-ubgoup, and even fuhe, a long a he ampling uni 16

17 wihin each level (e.g. po, hen plan wihin po, hen flowe wihin plan, ec. ae choen andomly. The eenial objecive of a need ANOVA i o diec he MSE of a yem ino i componen, heeby aceaining he ouce and magniude of eo in an expeimen o poce. Example of applicaion of need ANOVA ae: To aceain he magniude of eo a vaiou age of an expeimen o poce. To eimae he magniude of he vaiance aibuable o vaiou level of vaiaion in a udy of quaniaive geneic To dicove ouce of vaiaion in naual populaion in yemaic udie, ec. If you ae confued in a need deign, you can alway aveage he ubample and pefom a imple ANOVA. Thi i alo a good aegy o e if you need deign analyi i coec. The final P value will be he ame wih a coec need deign analyi and a non-need deign uing he aveage of he ubample. The advanage of doing he moe complex analyi including he ubample i o calculae he diffeen componen of vaiance Linea model fo ubampling Befoe we compue a need ANOVA, we hould examine he linea model upon which i i baed: Y ijk = + i + j(i + k(ij The inepeaion of,, and ae a befoe. Bu now wo andom elemen ae obained wih each obevaion: The j(i ae aumed nomal wih mean 0 and vaiance. The ubcip j(i indicae ha he j h level of eplicaion i need wihin he i h level of eamen. Noe: hi i a diffeen noaion fom ST&D. The j(i meaue, a befoe, he vaiaion among eal eplicaion wihin eamen goup, and he ubindex indicae ha hee ae ubample wihin he eplicaion. In he expeimen whee po ae andomized among eamen, and each po include 4 plan, j(i meaue he vaiaion among po mean (aveage of 4 plan wihin a eamen. The k(ij epeen he eo aociaed wih he vaiaion among ubample wihin an expeimenal uni. In he po expeimen k(ij meaue he vaiaion among he 4 plan wihin each po. The k(ij ae alo aumed nomal wih mean 0 and vaiance. Thi i epeened in he ample daa a: Y ijk =Y... + (Y i.. -Y... + (Y ij. - Y i.. + (Y ijk - Y ij. Remembe ha in hi noaion he do eplace a ubcip and indicae ha all value coveed by ha ubcip have been added Applying hi fomula o he po expeimen: 17

18 i meaue he diffeence beween a eamen mean and he oveall mean j(i meaue he diffeence beween a po mean and he mean of i aigned eamen. k(ij meaue he diffeence beween a plan and he mean of i po Need ANOVA wih equal ubample numbe: compuaion In hi expeimen, min plan ae expoed o ix diffeen combinaion of empeaue and dayligh and em gowh wa meaued a 1 week. The 6 eamen ae aigned andomly aco 18 po (i.e. 3 eplicaion pe eamen combinaion. Wihin each po ae fou plan (i.e. ubample. Someime we may be unceain a o whehe a faco i coed o need. If he level of he faco ae ju fo idenificaion (i no a claificaion cieia and can be enumbeed abiaily wihou affecing he analyi, hen he faco i need. Fo example, po 1,, 3 wihin eamen level 1 could be elabeled, 3, 1 wihou cauing any poblem. Tha i becaue po numbe i imply an ID, no a claificaion vaiable. Po 1 in eamen 1 ha nohing o do wih Po 1 in eamen. The daa (fom ST&D page 159: Teamen Low T, 8 h Low T, 1 h Low T, 16 h High T, 8 h High T, 1 h High T, 16 h Po numbe Po numbe Po numbe Po numbe Po numbe Po numbe Plan N o Po oal = Y ij T. oal = Y i T. mean= Y i In hi example, = 6, = 3, = numbe of ubample = 4, and n = = 7. Recall ha fo a CRD he um of quae aifie: i1 j1 i1 ( Y Y.. ( Y i. Y.. ( Y Y i. o TSS = SST + SSE. ij i1 j1 ij The degee of feedom aociaed wih hee um of quae ae n-1, -1, and n-, epecively. In he need deign, TSS and SST ae unchanged bu he SSE i paiioned ino wo componen. The euling equaion can be wien a: i1 j1 k 1 ( Y ijk Y... i1 ( Y i.. Y... i1 j1 ( Y ij. Y i.. i1 j1 k 1 ( Y ijk Y ij. o: TSS = SST + SSEE + SSSE 18

19 The wo eo em epeen he um of quae due o expeimenal eo and he um of quae due o ampling eo. In he po expeimen, SSEE epeen he vaiaion among po wihin eamen and SSSE epeen he vaiaion among plan wihin po. Need ANOVA able: Souce of vaiaion df SS MS F Expeced MS Teamen (τ i - 1 = 5 SST SST / 5 MST / MSEE /5 Exp. Eo (ε j(i ( - 1 = 1 SSEE SSEE / 1 MSEE / MSSE +4 Samp. Eo (δ k(ij n ( - 1 = 54 SSSE SSSE / 54 Toal n - 1 = 71 TSS In each cae, he numbe of degee of feedom i he poduc of he numbe of level aociaed wih each ubcip beween backe and he numbe of level minu one aociaed wih he ubcip ouide he backe. The expeced mean quae ae he heoeical model of he vaiance componen included in each MSE. The MSSE eimae (vaiaion among plan, and he MSEE eimae boh he vaiaion beween plan ( and he vaiaion beween po (. The la one i muliplied by 4 becaue po ae mean of 4 plan ( = /4 and o pu eveyhing in he ame cale ( i need o be muliplied. The eamen effec ae baed on eamen mean calculaed fom 1 plan (4*3, and ha i why i i muliplied by 1. The mo impoan pa of hi able: In eing a hypohei abou eamen mean, he appopiae divio fo F i he mean quae expeimenal eo (MSEE ince i include he vaiaion fom all ouce (po and plan ha conibue o he vaiabiliy of eamen mean excep he eamen effec hemelve. If you do no infom he aiical pogam he plan ae ubample, he pogam will auomaically divide by he MSSE, and he P value will anwe he queion: I hee a ignifican diffeence beween eamen o po? MST/MSSE->EMS= 4 +1 /5 inead of he one you hing you ae anweing which i: I hee a diffeen beween eamen? MST/MSEE->EMS= 1 /5 In a need deign he mo ciical pa i he elecion of he coec eo em 19

20 Eimaion of he diffeen vaiance componen in he po expeimen The main objecive in a need deign i o eimae he vaiance componen. To do hi, we deconuc he calculaed mean quae accoding o hei undelying heoeical model o expeced mean quae (EMS, la column in he able fo each componen of he linea model, a hown below: Vaiance Sum of Mean Vaiance Pecen of Souce df Squae Squae componen oal Toal % m % po % plan % MSSE =, o = 0.93 MSEE = +4, o = (MSEE - /4 = ( /4 = 0.30 MST = /5, o /5= (MST - MSEE/1 = ( /1 =.81 In hi example, he vaiaion among plan wihin a po i hee ime lage han he vaiaion among po wihin a eamen. In SAS, PROC VARCOMP compue hee vaiance componen fo diffeen model. Fo ou example expeimen hee: Poc GLM; Cla Tm Po; Model Gowh = Tm Po(Tm; Random Po(Tm; Te h = Tm e = Po(Tm; Poc Vacomp; Cla Tm Po; Model Gowh = Tm Po(Tm; Po(Tm: indicae ha po i a need faco in eamen. Po 1 in eamen 1 i no moe imila o po 1 in eamen han o po and 3. Random Po(Tm: Thi aemen ell SAS ha he po ae a andom faco (i.e. po 1, and 3 ae ju a andom ample, no a claificaion baed on a common popey. Te h = Tm e = Po(Tm: Thi aemen ell SAS which eo em o ue o e a paicula hypohei. Fo he hypohei abou eamen (h = Tm, he appopiae eo em i he MSEE (i.e. Po(m, o e = Po(Tm. Thi pecifie he e MST/MSEE. Noe ha you do no include a cla vaiable fo he la level of ub-ampling (in hi cae, plan. By defaul, SAS will ue hi la level of vaiaion (among plan wihin a po a he eo em fo he expeimen. Thi i why he e aemen i o impoan in a need deign: 0

21 SAS' defaul eo em (MSSE i coec fo eing diffeence among po wihin a eamen (MSEE, bu no fo eing he diffeence among eamen (MST. If you have wo level of neing (e.g. you meaue wo leave fom each plan, hen you include plan a a cla vaiable (bu no leaf and you indicae ha a plan i need in (po m. Poc GLM; Cla Tm Po Plan; Model Gowh = Tm Po(m Plan(Po Tm; Random Po(Tm Plan(Po Tm; Te h = Tm e = Po(Tm; Poc Vacomp; Cla Tm Po Plan; Model Gowh = Tm Po(m Plan(Po Tm; Opimal allocaion of eouce Addiional infomaion in Biomey Sokal & Rohlf page 309. One of he main eaon o ue a need deign i o inveigae how he vaiaion i diibued among expeimenal uni and among ubample (i.e. whee ae he ouce of eo in he expeimen. Once he vaiance componen of he expeimenal uni ( e.u. and he vaiance componen of he ubample ( ub ae known, he opimum numbe of ample and ubample can be calculaed uing he elaive co of expeimenal uni and ubample and he fomula below: To inoduce he idea of co, we wie a co funcion. Fo a wo-level need deign, he oal co (C will be he co of he ubample muliplied by he oal numbe of ubample (N ub plu he co of each expeimenal uni muliplied by he numbe of expeimenal uni (N eu : C N ub * N eu ( Cub N eu ( Ceu To find he numbe of ubample (N ub pe expeimenal uni ha will eul in imulaneou minimal co and minimal vaiance, he following fomula may be ued: N ub C C e. u. ub * * ub e. u. The opimum he numbe of ubample will inceae when he expeimenal uni ae moe expenive han he ubample, and when he ubample ae moe vaiable han he expeimenal uni. 1

22 If he co of ample and ubample i he ame, he opimum numbe of ubample in ou example can be calculaed a: ub 0.93 N ub 1.76 o plan pe po 0.30 e. u. In he cae of equal co he numbe of ubample i he quae oo of he aio beween ubample and ample vaiance. If he ubample ae 4 ime moe vaiable, he opimum i wo ubample. If he co i he ame and ub < e.u., i i bee o allocae all he eouce o expeimenal uni (in hi example, would be o ue one plan pe po. Theefoe, ubampling i only ueful when he vaiaion among ubample i lage han he vaiaion among expeimenal uni and/o he co of he ubample i malle han he co of he expeimenal uni.