4.1 Topc 4. Orthogonal contrasts [ST&D p. 183] ANOVA s a useful and powerful tool to copare several treatent eans. In coparng t treatents, the null hypothess tested s that the t true eans are all equal (H 0 : µ 1 = µ 2 =... = µ t ). If the F test s sgnfcant, one accepts the alternatve hypothess, whch erely states that they are not all equal (.e. at least one ean s dfferent). Snce the test does not tell you whch ean(s) s/are dfferent, the nforaton provded by ths ntal F test s lted. Further coparsons to deterne whch specfc treatent eans are dfferent can be carred out by further parttonng the treatent su of squares (SST) to provde addtonal F tests to answer planned questons. The orthogonal contrast approach to ean separaton s descrbed as planned F tests. These tests are planned n the sense that the questons of nterest are decded upon before lookng at the data. In fact, the questons of nterest dctate the treatent structure fro the very begnnng of the experent. Wth planned tests, therefore, there s a pror knowledge, based ether on bologcal consderatons or on the results of prelnary nvestgatons, as to whch coparsons are ost portant to ake. Sad another way, f the nvestgator has specfc questons to be answered, treatents are chosen (.e. a treatent structure s desgned) to provde nforaton and statstcal tests to answer those specfc questons. An experenced nvestgator wll select treatents so that the treatent su of squares (SST) can be parttoned perfectly to answer as any ndependent (.e. orthogonal) questons as there are degrees of freedo for treatents n the ANOVA. Consequently, another nae of these tests s sngle degree of freedo tests. 4. 1. Defntons of contrast and orthogonalty [ST&D p. 183] "Contrast" s atheatcal jargon for a lnear cobnaton of ters (a polynoal) whose coeffcents su to zero. In ANOVA, "contrast" assues a ore lted defnton: A contrast (Q) s a lnear cobnaton of two or ore factor level eans whose coeffcents su to zero: Q 1 c Y, wth the constrant that c 0 As an exaple, consder a coparson of two eans, the splest contrast: 1 2 Ths s the sae as 0, where c 1, c 1, and c c 0 1 2 1 1 2 1 2 It s essental that the su of the coeffcents for each coparson s zero. The ters are the treatent eans (t can also be the treatent sus), and t, s Y.
4.2 the nuber of treatents beng copared. For convenence, the c s are usually ntegers. A contrast always has a sngle degree of freedo. Orthogonal: Now consder two contrasts Suppose Qc cy.; and Qd d Y.. 1 1 These two contrasts are sad to be orthogonal to one another f the su of the products of any par of correspondng coeffcents s zero. 1 c d 0 ( or 1 c d / n 0 for unbalanced desgns) So, orthogonalty s a property of a set of two contrasts. A set of ore than two contrasts s sad to be orthogonal only f each and every par wthn the set exhbts par-wse orthogonalty, as defned above. To declare a set of four contrasts (Q 1 Q 4 ) to be orthogonal, therefore, one ust show that each of the sx possble pars are orthogonal: (Q 1 and Q 2, Q 1 and Q 3, Q 1 and Q 4, Q 2 and Q 3, Q 2 and Q 4, Q 3 and Q 4 ). Why do we care? Reeber n our dscusson of CRDs that the total SS could be perfectly parttoned nto two parts: The treatent SS and the error SS: TSS = SST + SSE Ths perfect parttonng s possble due to the fact that, atheatcally, SST and SSE are orthogonal to one another. In an analogous way, orthogonal contrasts allow us to partton SST (.e. decopose the relatvely unnforatve H 0 of the ANOVA) nto a axu of (t - 1) eanngful and targeted coparsons nvolvng dfferent cobnatons of eans. Suppose a set of (t 1) contrasts s orthogonal. If the SS for each contrast are added together, ther su wll exactly equal the SST for the orgnal experent. Ths eans that an experent can be parttoned nto (t 1) separate, ndependent experents, one for each contrast. Exaple Suppose we are testng three treatents, T1, T2 and T3 (control) wth treatent eans 1, 2, and 3 (two d.f.). The null hypothess for the ANOVA s H 0 : 1 = 2 = 3 whch uses both degrees of freedo n one test (df t = t-1 = 2). Snce there are two degrees of freedo for treatents, there are n prncple two ndependent coparsons that can be ade.
4.3 For exaple, one could n prncple test the two hypotheses that 1 and 2 are not sgnfcantly dfferent fro the control: 1 = 3 and 2 = 3. As usual, we represent the eans wth ther estates Y 1.- 1 = 3 can be rewrtten as 1 1 + 0 2-1 3 = 0 the coeffcents of ths contrast are: c 1 = 1, c 2 = 0, c 3 = -1 2.- 2 = 3 can be rewrtten as 0 1 + 1 2-1 3 = 0 the coeffcents of ths contrast are: d 1 = 0, d 2 = 1, d 3 = -1 These lnear cobnatons of eans are contrast because 1 c 0 (1 + 0 + (-1) = 0); and d 0 (0 + 1 + (-1) = 0). 1 However these contrasts are not orthogonal because: 1 c d 0 (c 1 d 1 + c 2 d 2 + c 3 d 3 = 0 + 0 + 1 = 1). So, not every par of hypotheses can be tested usng ths approach. In addton to sung to 0, the c coeffcents are alost always taken to be ntegers, a constrant whch severely restrcts ther possble values. For t = 3 such a set of values are 1) c 1 = 1, c 2 = 1, c 3 = -2; 2) d 1 = 1, d 2 = -1, d 3 = 0. These are contrasts snce (1 + 1 + (-2) = 0 and 1+ (-1) + 0= 0) and are orthogonal because (c 1 d 1 + c 2 d 2 + c 3 d 3 = 1 + (-1) + 0 = 0). Just as not all sets of hypotheses can be asked usng orthogonal contrasts, not all sets of orthogonal contrasts correspond to eanngful (or nterestng) hypotheses. In ths exaple, the contrasts are, n fact, nterestng. The hypotheses they defne are: 1) The average of the two treatents s equal to the control (.e. s there a sgnfcant average treatent effect?); and 2) The two treatents are equal to one another (.e. s one treatent sgnfcantly dfferent fro the other?).
4.4 There are two general knds of lnear cobnatons: Class coparsons Trend coparsons. 4. 2. Class coparsons The frst category of hypothess we can pose usng orthogonal contrasts s class (or group) coparsons. Such contrasts copare specfc treatent eans or cobnatons of treatent eans, grouped s soe eanngful way. The procedure s llustrated by an exaple on page 185 of ST&D, whch nvolves the nt data dscussed n Topc 3. To llustrate orthogonal contrasts n class coparsons we wll use the data gven n Table 4.1. The analyss of varance for ths experent s gven n Tables 4.2. Table 4.1. Results (g shoot dry weght) of an experent (CRD) to deterne the effect of seed treatent by acds on the early growth of rce seedlngs. Treatent Replcatons Total Mean Y. Y. Control 4.23 4.38 4.1o 3.99 4.25 20.95 4.19 HC1 3.85 3.78 3.91 3.94 3.86 19.34 3.87 Proponc 3.75 3.65 3.82 3.69 3.73 18.64 3.73 Butyrc 3.66 3.67 3.62 3.54 3.71 18.20 3.64 Overall Table 4.2. ANOVA of data n Table 4.1 (t = 4, r = 5) Y.. = 77.13 Y.. = 3.86 Source of Su of Varaton df Squares Total 19 1.0113 Mean Squares F Treatent 3 0.8738 0.2912 33.87 Error 16 0.1376 0.0086 The treatent structure of ths experent suggests that the nvestgator had several specfc questons n nd fro the begnnng: 1) Do acd treatents affect seedlng growth? 2) Is the effect of =organc acds dfferent fro that of norganc acds? 3) Is there a dfference n the effects of the two dfferent organc acds?
4.5 These are planned questons and so are approprate canddates for posng va orthogonal contrasts. To do ths, we ust restate these questons atheatcally, as lnear cobnatons of treatents. In the followng table (Table 4.3), coeffcents are shown that translate these three planned questons nto contrasts. Table 4.3. Orthogonal coeffcents for parttonng the SST of Table 4.2. nto 3 ndependent tests. Control HC1 Proponc Butyrc Control vs. acd +3-1 -1-1 Inorganc vs. organc 0-2 +1 +1 Between organcs 0 0 +1-1 Totals 20.95 19.34 18.64 18.2 Coparsons Means 4.19 3.87 3.73 3.64 The 1 st contrast (frst row) copares the control group to the average of the three acd-treated groups, as can be seen fro the followng anpulatons: 3μ Cont - 1μ HCl - 1μ Prop - 1μ But = 0 μ Cont = (1/3)*(1μ HCl + 1μ Prop + 1μ But ) Mean of the control group = Mean of all acd-treated groups The H 0 for ths 1 st contrast s that there s no average effect of acd treatent on seedlng growth. Snce ths H o nvolves only two group eans, t costs 1 df. The 2 nd contrast (second row) copares the norganc acd group to the average of the two organc acd groups: 0μ Cont + 2μ HCl - 1μ Prop - 1μ But = 0 μ HCl = (1/2)*(1μ Prop + 1μ But ) Mean of the HCl group = Mean of all organc acd-treated groups The H 0 for ths second contrast s that the effect of the norganc acd treatent on seedlng growth s no dfferent fro the average effect of organc acd treatent. Snce ths null hypothess nvolves only two group eans (eans than before), t also costs 1 df. Fnally, the thrd contrast (thrd row of coeffcents) copares the two organc acd groups to each other: 0μ Cont + 0μ HCl + 1μ Prop - 1μ But = 0 1μ Prop = 1μ But
4.6 The H 0 for ths thrd contrast s that the effect of the proponc acd treatent on seedlng growth s no dfferent fro the effect of butyrc acd treatent. Snce ths null hypothess nvolves only two group eans (dfferent eans than before), t also costs 1 df. At ths pont, we have spent all our avalable degrees of freedo (df trt = t 1 = 4 1 = 3). Because each of these questons are contrasts (each row of coeffcents sus to zero) and because the set of three questons s orthogonal (verfy ths for yourself), these three queston perfectly partton SST nto three coponents, each wth 1 df. The SS assocated wth each of these contrasts serve as the nuerators for three separate F tests, one for each coparson. The crtcal F values for these sngle df tests are based on 1 df n the nuerator and df Error n the denonator. All of ths can be seen n the expanded ANOVA table below. Table 4.4 Orthogonal parttonng of SST va contrasts. Source df SS MS F Total 19 1.0113 Treatent 3 0.8738 0.2912 33.87 1. Control vs. acd 1 0.7415 0.7415 86.22 2. Inorg. vs. Org. 1 0.1129 0.1129 13.13 3. Between Org. 1 0.0194 0.0194 2.26 Error 16 0.1376 0.0086 Notce that SST = SS Contrast1 + SS Contrast2 + SS Contrast3. Ths perfect parttonng of SST aong ts degrees of freedo s a drect consequence of the orthogonalty of the posed contrasts. When coparsons are not orthogonal, the SS for one coparson ay contan (or be contaned by) part of the SS of another coparson. Therefore, the concluson fro one test ay be nfluenced (or contanated) by another test and the SS of those ndvdual coparsons wll not su to SST. Coputaton: The coputaton of the su of squares for a sngle degree of freedo F test for lnear cobnatons of treatent eans s ( ( 2 cy.) SS ( Q) MS( Q) 2 c / r ) 2 ( cy.) Ths expresson splfes to n balanced desgns (all r's equal) 2 ( c ) / r SS 1 (control vs. acd) = [3(4.19) 3.64 3.73 3.87] 2 / [(12)/5] = 0.74 SS 2 (norg. vs. org.) = [3.64 + 3.73 2(3.87)] 2 / [(6)/5] = 0.11 SS 3 (between org.) = [-3.64 + 3.73] 2 / [(2)/5] = 0.02
4.7 Note: ST&D forulas for contrasts (p. 184) are for treatent totals nstead of treatent eans. The treatents eans forula s requred for unbalanced desgns. Fro ths analyss, we conclude that n ths experent acds sgnfcantly reduce seedlng growth (F = 86.22, p < 0.01), that the organc acds cause sgnfcantly ore reducton than the norganc acd (F = 13.13, p < 0.01), and that the dfference between the organc acds s not sgnfcant (F = 2.26, p > 0.05). Constructon of coeffcents for class coparsons (Lttle & Hlls p 66). Contrast coeffcents for a class coparson can always be deterned by wrtng the null hypothess n atheatcal for, ovng all ters to the sae sde of the equaton, and ultplyng by whatever factor s needed to turn the coeffcents nto ntegers. Ths s a general strategy. What follows are ore recpe-lke, step-bystep operatons to arrve at the sae results: 1. When the two groups of eans beng copared each contan the sae nuber of treatents assgn +1 to the ebers of one group and -1 to the ebers of the other. Thus for lne 3 n Table 4.3, we are coparng two eans, and assgn coeffcents of 1 (of opposte sgn) to each. The sae procedure extends to the case of ore than one treatent per group. 2. When coparng groups contanng dfferent nubers of treatents, assgn to the frst group coeffcents equal to the nuber of treatents n the second group; to the second group, assgn coeffcents of opposte sgn, equal to the nuber of treatents n the frst group. Thus, f aong 5 treatents, the frst two are to be copared to the last three, the coeffcents would be +3, +3, -2, -2, -2. In Table 4.3, where the control ean s copared wth the ean of the three acds, we assgn a 3 to the control and a 1 to each of the three acds. Opposte sgns are then assgned to the two groups. It s ateral as to whch group gets the postve or negatve sgn snce t s the su of squares of the coparson that s used n the F-test. 3. The coeffcents for any coparson should be reduced to the sallest possble ntegers for each calculaton. Thus, +4, +4, -2, -2, -2, -2, should be reduced to +2, +2, -1, -1, -1, -1. 4. The coeffcents for an nteracton coparson are deterned by sply ultplyng the correspondng coeffcents of the two underlyng an coparsons. See next exaple (Table 4.5).
4.8 Exaple: Fertlzer experent desgned as a CRD wth four treatents. The four treatents result fro all possble cobnatons of two levels of both ntrogen (N 0 = 0 N, N 1 = 100 lbs N/ac) and phosphorus (P 0 = 0 P, P 1 = 20 lbs P/ac). The questons ntended by ths treatent structure are: 1. Is there an effect of N on yeld? 2. Is there an effect of P on yeld? 3. Is there an nteracton between N and P on yeld? Equvalent ways of statng the nteracton queston: a) Is the effect of N the sae n the presence of P as t s n the absence of P? b) Is the effect of P the sae n the presence of N as t s n the absence of N?) The followng table (Table 4.5) presents the contrast coeffcents for these planned questons. Table 4.5. Fertlzer experent wth 4 treatents, 2 levels of N and 2 of P. N 0 P 0 N 0 P 1 N 1 P 0 N 1 P 1 Between N -1-1 1 1 Between P -1 1-1 1 Interacton (NxP) 1-1 -1 1 The coeffcents for the 1 st two coparsons are derved by rule 1. The nteracton coeffcents are the result of ultplyng the coeffcents of the frst two lnes. Note that the su of the coeffcents of each coparson s 0 and that the su of the cross products of any two coparsons s also 0. When these 2 condtons are et, the coparsons are sad to be orthogonal. Ths ples that the concluson drawn for one coparson s ndependent of (not nfluenced by) the others. 4. 3. Trend coparsons Experents are often desgned to characterze the effect of ncreasng levels of a factor (e.g. ncreents of a fertlzer, plantng dates, doses of a checal, concentratons of a feed addtve, etc.) on soe response varable (e.g. yeld, dsease severty, growth, etc.). In these stuatons, the experenter s nterested n the dose response relatonshp. Such an analyss s concerned wth overall trends and not wth parwse coparsons. The splest exaple nvolves a sngle factor wth three levels. Ths s very coon stuaton n genetc experents, where the levels are: 0 dose of allele A n hoozygous BB ndvduals
4.9 1 dose of allele A n heterozygous AB ndvduals 2 doses of allele A n hoozygous AA ndvduals. Wth the use of olecular arkers t s now easy to score the genotype of the ndvduals of a segregatng populaton and classfy the nto one of these three groups (AA, AB, BB). These ndvduals are also phenotyped for the trat of nterest. Suppose 40 segregatng F 2 ndvduals genotyped for a certan olecular known to be lnked to a gene affectng N translocaton to the gran. The Ntrogen content of the seed s easured n the sae 40 ndvduals. Table 4.6. Genetc exaple of orthogonal contrasts. Ntrogen content (g) of seeds of three dfferent genotypes. Genotype (BB) Genotype (AB) Genotype (AA) 0 doses, A allele 1 dose, A allele 2 doses, A allele 12.0 13.5 13.8 12.5 13.8 14.5 12.1 13.0 13.9 11.8 13.2 14.2 12.6 13.0 14.1 12.8 12.9 13.4 12.7 13.6 Unequal replcaton s coon n genetc experent due to segregaton ratos. In F 2 populatons, the expected rato of hoozygous to heterozygous ndvduals s 1:2:1, whch s what we see n the dataset above. Each ndvdual s an ndependent replcaton of ts respectve genotype; so there are fve replcatons of genotype BB, ten replcatons of genotype AB, and fve replcatons of genotype AA n ths experent. The "treatent" s dosage of the A allele, and the response varable s seed ntrogen content. Wth three levels of dosage, the ost coplcated response the data can reveal s a quadratc relatonshp between dosage (D) and N content: N = ad 2 + bd + c Ths quadratc relatonshp s coprsed of two coponents: A lnear coponent (slope b), and a quadratc coponent (curvature a). It just so happens that, wth 2 treatent degrees of freedo (df trt = t 1 = 3 1 = 2), we can construct orthogonal contrasts to probe each of these coponents. To test the hypothess that b = 0 (.e. there s 0 slope n the overall dosage response relatonshp), we choose H 0 : μ BB = μ AA. If the eans of the two extree dosages are equal, b = 0. As a contrast, ths H 0 takes the for: 1μ BB + 0μ AB - 1μ AA = 0.
4.10 To test the hypothess that a = 0 (.e. there s zero curvature n the overall dosage response relatonshp), we choose H 0 : μ AB = (1/2)*(μ AA + μ BB ). Because the dosage levels are equally spaced (0, 1, 2), a perfectly lnear relatonshp (.e. zero curvature) would requre that the average of the extree dosage levels [(1/2)*(μ AA + μ BB )] exactly equal the ean of the heterozygous group (μ AB ). As a contrast, ths H 0 takes the for: 1μ BB - 2μ AB + 1μ AA = 0. A quck nspecton shows each of these polynoals to be contrasts (.e. ther coeffcents su to zero) as well as orthogonal to each other (1*1 + 0*(-2) + (-1)*1 = 0). Constructng F tests for these contrasts follows the exact sae procedure we saw above n the case of class coparsons. So ths te, let's use SAS: SAS progra: Data GeneDose; Input Genotype $ N; Cards; BB 12.0 BB 12.5 BB 12.1 BB 11.8 BB 12.6 AB 13.5 AB 13.8 AB 13.0 AB 13.2 AB 13.0 AB 12.8 AB 12.9 AB 13.4 AB 12.7 AB 13.6 AA 13.8 AA 14.5 AA 13.9 AA 14.2 AA 14.1 ; Proc GLM Order = Data; Class Genotype; Model N = Genotype; Contrast 'Lnear' Genotype 1 0-1; Contrast 'Quadratc' Genotype 1-2 1; Run; Qut;
4.11 The resultant ANOVA table: Source df SS MS F p Total 19 11.022 Model 2 9.033 4.5165 38.60 < 0.0001 Error 17 1.989 0.117 R-Square: 0.819543 Contrast df SS MS F p Lnear 1 9.025 9.025 77.14 < 0.0001 Quadratc 1 0.008 0.008 0.07 0.7969 The fact that the contrast SS su perfectly to the SST s a verfcaton of ther orthogonalty. The sgnfcant lnear contrast (p < 0.0001) leads us to the reject ts H 0. There does appear to be a sgnfcant, nonzero lnear coponent to the response. The non-sgnfcant quadratc contrast (p = 0.7969), however, leads us not to reject ts H 0. Snce the quadratc contrast s not sgnfcant we do not reject the hypothess of a lnear response. Fro the boxplot graph, there does not appear to be a quadratc coponent to the response (the ddle ean s algned wth the lne between the extree eans): μ 1 - μ 2 We would conclude fro all ths that the dosage response of ntrogen seed content to the presence of allele A s lnear. Before we ove on, notce that when there s no sgnfcant quadratc response, the F value of the lnear response (77.14, crtcal value F 2,17 = 3.59) s twce as large as the Model F value (38.50, crtcal value F 1,17 = 4.45). The reason for ths: In the lnear contrast, MS = SS/1, whle n the coplete Model, MS = SS/2 (.e. the full SST s dvded n half and arbtrarly assgned equally to both effects). When a quanttatve factor exhbtng a sgnfcant lnear dose response s easured at several levels, t s possble to have a non-sgnfcant overall treatent F test but a sgnfcant lnear response. Ths s because the overall treatent F test dvdes the full SST equally across any effects, ost of whch are non-sgnfcant. Ths obscures the sgnfcance of the true lnear effect. In such cases, contrasts sgnfcantly ncrease the power of the test.
4.12 Here s a slar dataset, but now the response varable s days to flowerng (DTF). Table 4.7 Days to flowerng (DTF) of seeds of three dfferent genotypes. Genotype (BB) Genotype (AB) Genotype (AA) 0 doses, A allele 1 dose, A allele 2 doses, A allele 58 71 73 51 75 68 57 69 70 59 72 71 60 68 67 73 69 70 71 72 The SAS codng s dentcal n ths case. The resultant ANOVA table: Source df SS MS F p Total 19 811.20 Model 2 698.40 349.200 52.63 < 0.0001 Error 17 112.80 6.635 R-Square: 0.860947 Contrast df SS MS F p Lnear 1 409.6 409.6 61.73 < 0.0001 Quadratc 1 288.8 288.8 43.52 < 0.0001 Agan, the contrast SS su perfectly to the SST, a verfcaton of ther orthogonalty. The sgnfcant lnear contrast (p < 0.0001) leads us to the reject ts H 0. There does appear to be a sgnfcant, nonzero lnear coponent to the response. And the sgnfcant quadratc contrast (p < 0.0001), leads us to reject ts H 0 as well. There does appear to be a sgnfcant, nonzero quadratc coponent to the response. All ths can be seen qute easly n the followng cobned boxplot: μ 2 - (1/2)*(μ 1 + μ 3 ) 0
4.13 We would conclude fro all ths that the dosage response of flowerng to the presence of allele A has both a lnear and a quadratc coponent. In genetc ters, there s donance. If we were to analyze ths last exaple va a sple lneal regresson, we would obtan the followng results: Source df SS MS F p Total 19 811.20 Model 1 409.60 409.6 18.36 0.0004 Error 18 401.60 22.3 401.6 = 112.8 + 288.8 The F value s saller (18.36 < 61.73) because the quadratc SS (288.8) s now ncluded n the error su of squares (401.6 = 112.8 + 288.8). The essage: An ANOVA wth lnear and quadratc contrasts s ore senstve to lnear effects than a lnear regresson test. A quadratc regresson test, however, wll yeld dentcal results to our analyss usng contrasts. Coeffcents for trend coparsons The c coeffcents used for trend coparsons (lnear, quadratc, cubc, quartc, etc.) aong equally spaced treatents are lsted below, taken fro Table 15.12 (ST&D 390). Contrast coeffcents for trend coparsons for equally spaced treatents No. of Treat. Response c1 c2 c3 c4 c5 c6 2 Lnear -1 1 3 4 5 6 Lnear -1 0 1 Quadratc 1-2 1 Lnear -3-1 1 3 Quadratc 1-1 -1 1 Cubc -1 3-3 1 Lnear -2-1 0 1 2 Quadratc 2-1 -2-1 2 Cubc -1 2 0-2 1 Quartc 1-4 6-4 1 Lnear -5-3 -1 1 3 5 Quadratc 5-1 -4-4 -1 5 Cubc -5 7 4-4 -7 5 Quartc 1-3 2 2-3 1 Quntc -1 5-10 10-5 1
4.14 As argued n the two prevous exaples, the values of these coeffcents ultately can be traced back to sple geoetrc arguents. The plot below shows the values of the lnear and quadratc coeffcents for t = 5: X +2 X X X T1 X X T5 X -2 X To llustrate the procedure for evaluatng a trend response when ore treatent levels are nvolved, we wll use the data fro Table 15.11 (ST&D 387). To splfy atters, we wll treat the blocks sply as replcatons n a CRD. Table 4. 8. Parttonng SST usng orthogonal polynoals. Yeld of Ottawa Mandarn soybeans grown n MN (bushels / acre). [ST&D 387] Rep.* Row spacng (n nches) 18 24 30 36 42 1 33.6 31.1 33 28.4 31.4 2 37.1 34.5 29.5 29.9 28.3 3 34.1 30.5 29.2 31.6 28.9 4 34.6 32.7 30.7 32.3 28.6 5 35.4 30.7 30.7 28.1 29.6 6 36.1 30.3 27.9 26.9 33.4 Means 31.15 31.63 30.17 29.53 30.03 * Blocks treated as replcatons n ths exaple Frst of all, note that the treatent levels are equally spaced (18, 24, 30, 36, 42 6 nches between adjacent levels). Trend analyss va contrasts s greatly splfed when treatent levels are equally spaced (ether arthetc or log scales). The contrast coeffcents and the analyss: Y. Row spacng c 2 c 2 / r 18 24 30 36 42 SS F Means 35.15 31.63 30.17 29.53 30.03 Lnear -2-1 0 1 2 152.11 1.67 91.27 28.8** * Quadratc 2-1 -2-1 2 78.62 2.33 33.69 10.6** Cubc -1 2 0-2 1 0.84 1.67 0.50 0.16 NS
4.15 Quartc 1-4 6-4 1 2.30 11.67 0.20 In ths trend analyss, we perfectly parttoned SST (125.66) aong the four degrees of freedo for treatent, each degree of freedo correspondng to an ndependent, orthogonal contrast (lnear, quadratc, cubc, and quartc coponents to the overall response). We conclude fro ths analyss that the relatonshp between row spacng and yeld has sgnfcant lnear and sgnfcant quadratc coponents. The cubc and quartc coponents are not sgnfcant. Ths can be seen easly n a scatterplot of the data: 0.06 NS Unequally spaced treatents There are equatons to calculate coeffcent slar to those of Table 15.12 for unequally spaced treatent levels and unequal nubers of replcatons. The ablty to copute such sus of squares usng orthogonal contrasts was crucal n the days before coputers. But now t s easer to pleent a regresson approach, whch does not requre equal spacng between treatent levels [ST&D 388]. The SAS code for a full regresson analyss of the soybean yeld data: Data Rows; Do Rep = 1 to 6; Do Sp = 18,24,30,36,42; Input Yeld @@; Output; End; End; Cards; 33.6 31.1 33 28.4 31.4 37.1 34.5 29.5 29.9 28.3 34.1 30.5 29.2 31.6 28.9 34.6 32.7 30.7 32.3 28.6 35.4 30.7 30.7 28.1 29.6 36.1 30.3 27.9 26.9 33.4 ; Proc GLM Order = Data; Model Yeld = Sp Sp*Sp Sp*Sp*Sp Sp*Sp*Sp*Sp; Run; Qut; Note the absence of a Class stateent! In regresson, we are not nterested n the ndvdual levels of the explanatory varable; we are nterested n the nature of overall trend. The output:
4.16 Su of Source DF Squares Mean Square F Value Pr > F Model 4 125.6613333 31.4153333 9.90 <.0001 Error 25 79.3283333 3.1731333 Corrected Total 29 204.9896667 R-Square Coeff Var Root MSE Yeld Mean 0.613013 5.690541 1.781329 31.30333 Source DF Type I SS Mean Square F Value Pr > F Sp 1 91.26666667 91.26666667 28.76 <.0001 *** Sp*Sp 1 33.69333333 33.69333333 10.62 0.0032 ** Sp*Sp*Sp 1 0.50416667 0.50416667 0.16 0.6936 NS Sp*Sp*Sp*Sp 1 0.19716667 0.19716667 0.06 0.8052 NS Ths s the exact result we obtaned usng contrasts (see code and output below): Data Rows; Do Rep = 1 to 6; Do Sp = 18,24,30,36,42; Input Yeld @@; Output; End; End; Cards; 33.6 31.1 33 28.4 31.4 37.1 34.5 29.5 29.9 28.3 34.1 30.5 29.2 31.6 28.9 34.6 32.7 30.7 32.3 28.6 35.4 30.7 30.7 28.1 29.6 36.1 30.3 27.9 26.9 33.4 ; Proc GLM Order = Data; Class Sp; Model Yeld = Sp; Contrast 'Lnear' Sp -2-1 0 1 2; Contrast 'Quadratc' Sp 2-1 -2-1 2; Contrast 'Cubc' Sp -1 2 0-2 1; Contrast 'Quartc' Sp 1-4 6-4 1; Run; Qut; Su of Source DF Squares Mean Square F Value Pr > F Model 4 125.6613333 31.4153333 9.90 <.0001 Error 25 79.3283333 3.1731333 Corrected Total 29 204.9896667 R-Square Coeff Var Root MSE Yeld Mean 0.613013 5.690541 1.781329 31.30333 Source DF Type III SS Mean Square F Value Pr > F Sp 4 125.6613333 31.4153333 9.90 <.0001 Contrast DF Contrast SS Mean Square F Value Pr > F Lnear 1 91.26666667 91.26666667 28.76 <.0001 *** Quadratc 1 33.69333333 33.69333333 10.62 0.0032 ** Cubc 1 0.50416667 0.50416667 0.16 0.6936 NS Quartc 1 0.19716667 0.19716667 0.06 0.8052 NS So, as long as the treatent levels are equally spaced, the results are the sae for both analyses. The ultple regresson analyss can be used wth unequally spaced
4.17 treatents, but the orthogonal contrast analyss, wth the provded coeffcents, cannot. Soe rearks on treatent levels for trend analyss The selecton of dose levels for a ateral depends on the objectves of the experent. If t s known that a certan response s lnear over a gven dose range and one s only nterested n the rate of change, two doses wll suffce, one low and one hgh. However, wth only two doses there s no nforaton avalable to verfy the ntal assupton of lnearty. It s good practce to use one extra level so that devaton fro lnearty can be estated and tested. Slarly, f a quadratc response s expected, a nu of four dose levels are requred to test whether or not a quadratc odel s approprate. The varablty n agrcultural data s generally greater than for physcal and checal laboratory studes, as the experental unts are subject to less controllable envronental nfluences. These varatons cause dffculty n analyzng and nterpretng cobned experents that are conducted over dfferent years or across dfferent locatons. Furtherore, true response odels are rarely known. For these reasons, agrcultural experents usually requre four to sx levels to characterze a dose-response curve. Fnal coent about orthogonal contrasts: Powerful as they are, contrasts are not always approprate. If you have to choose, eanngful hypotheses are ore desrable than orthogonal ones!