NON-PARAMETRIC METHODS IN ANALYSIS OF EXPERIMENTAL DATA

NON-PARAMETRIC METHODS IN ANALYSIS OF EXPERIMENTAL DATA Rajender Parsad I.A.S.R.I., Library Aenue, New Delhi 0 0. Introduction In conentional setup the analysis of experimental data is based on the assumptions like normality, independence and constant ariance of the obserations. Howeer, there may arise experimental situations where these assumptions, particularly the assumption of normality, may not be satisfied. In such situations non-parametric test procedures may become quite useful. A lot of attention is being made to deelop non-parametric tests for analysis of experimental data. Most of these non-parametric test procedures are based on rank statistic. The rank statistic has been used in deelopment of these tests as the statistic based on ranks is. distribution free. easy to calculate and 3. simple to explain and understand. The another reason of use of the rank statistic is due to the well known result that the aerage rank approaches normality quickly as n (number of obserations) increases, under the rather general conditions, while the same might not be true for the original data {see e.g. Conoer and Iman (976)}. The non-parametric test procedures aailable in literature coer completely randomized designs, randomized complete block designs, balanced incomplete block designs, design for bioassays, split plot designs, cross-oer designs and so on. For excellent and elaborate discussions on non-parametric tests in the analysis of experimental data, one may refer to Sen (996), Deshpande, Gore and Shanubhogue (995) and Hollander and Wolfe (999). In the present talk we shall concentrate on the non-parametric test procedure for analysis of one-way and two-way classified data. Some of these procedures are:. Kruskal-Wallis Test. Friedman Test 3. Durbin Test 4. Skillings and Mack Test 5. Gore test for Multiple Obserations per Plot In the sequel we describe each of these tests in brief.. Non-parametric Procedures for the Analysis of Experimental Data In this section, we shall describe the non-parametric tests for analysis of experimental data generated through completely randomized designs (CRD) and block designs. We begin with the test procedure for the experimental data generated through CRD

.. Kruskal Wallis Test This is used for testing the equality of seeral independent samples. Hence, this test is quite useful for the analysis of experimental data generated through a completely randomized design. Consider that there are treatments and i th treatment is replicated r i times; i =,,,, such that N = r i. The data generated through CRD can be represented by usual one-way classified linear model as y ij = µ + τ i + ij i =,,, ; j =,,, r i. where y ij is the yield or response of j th replication of i th treatment, µ is the general mean, τ i is the i th treatment effect, ij is the error due to i th treatment and j th obseration. The experimenter is interested to test the equality of treatments effects against the alternatie that at least two of the treatments effects are not equal. In other words, we want to test the null hypothesis H 0 : τ = τ = τ 3 = = τ =τ (say) against the alternatie H : at least two of the τ i s are different. To test the aboe null hypothesis using the Kruskal-Wallis Statistic, we rank all N obserations by giing rank to smallest obseration and N to largest obseration. Once this is done, we obtain the sum and aerage of the ranks of the obserations pertaining to each of the treatments. Now, if the treatment effects are equal, then the aerage ranks expected to be same, the differences if any, are due to sampling fluctuations. The Kruskal-Wallis statistic is based on the assessment of the differences among the aerage ranks. This may be explained as below: Let R ij be the rank of y ij ; i =,, 3,, ; j =,,, r i and R i = ranks of the obserations pertaining to the i th treatment) and i i r i R ij j= i ( the sum of the R = R / r (the aerage of the ranks of the obserations pertaining to the i th treatment). Let R is the mean of the all Kruskal-Wallis statistic is, then, gien by T = N ri ( N + ) ( R R ) R i N + T = ri N( N + ) ri i R i. The = Ri N( N + ) ri 3( N + ) The method of determining the significance of the obsered alues of T depends on the number of treatments () and their replications (r i ). The Null Distribution (distribution under 680

null hypothesis) of T for =3 and r i 5 is extensiely tabulated and aailable in seeral texts. For a ready reference, the Table has been included in the Appendix-I of the present text. In other cases, under null hypothesis, T may be approximated by the χ with (-) degree of freedom. Remark.: When ties occur between two or more obserations, each obseration is gien the mean of ranks for which it is tied. The T-statistic after correcting the effect of ties is computed by using following formula T = N( N + ) g ( ts s= 3 R i 3( N + ) r i 3 ts ) /( N N ) where g is number of treatments of different tied ranks, t s is the number of tied ranks in the s th treatment with tied ranks. The effect of correcting for ties is to increase the alue of T and thus to make the result more significant than it would hae been had no correction been made. Therefore, if one is able to reject null hypothesis without making the correction, one will be able to reject H 0 at an een more stringent leel of significance if the correction is used. The Kruskal-Wallis statistic may be deried by taking the help of Mann - Whitney Wilcoxon test for two independent sample that correspond to treatments that are replicated r and r times respectiely. Some steps of the deriation are listed as below: Suppose,, y,,,,, y,, y y 3 y y r y 3 y be two random samples of size r r and r respectiely such that r +r = N. Let us order the combined sample from smallest (rank ) to largest (rank r +r ) and let R( y ) = rank of y in the combined sample, k =,,, r. k k Then R( yk ) r+r for k =,,, r. Under H 0 we expect the r ranks of the y s namely R( ), R( ),, R( y ) to be randomly spread out among the ranks,,, r+r. y y r The test based on the rank sum statistic r T = R( yk ) k= Now we assumed that the sample with fewer obserations is the first sample (y ) so that r r. Next note that the least alue of y corresponds to the ranks,,, r and the maximum alue corresponds to ranks +, +,, + r. It follows that r r r 68

r r ( r + ) r ( r + r + ) y ( r + l ) = l= Finally, under H 0 : T has a symmetric distribution about its mean r ( r + r +)/. It follow that a left tail cumulatie probability equals the corresponding right tail cumulatie probability. For larger alue of r or r we use the fact that r ( r + r + ) E(T ) = and Then for large r + r, we know that rr ( r + r Var(T ) = + ) Z = R r ( r r r ( r + r + r + ) / + ) / has approximately a standard normal distribution, so that Z r ( r + r + ) = R rr ( r r ) + + r R r r + = + r ( r + r + ) r has approximately a χ with one degree of freedom. The first thing to note is that since R = [N(N+)]/ R, then after simplification we can write R + N + R + N Z = r r N( N + ) r r after generalization this equation in term of ( > ) treatments assuming that i th treatment is replicated r i times such that aboe. r i = N ~ χ ( ) then we get T statistic gien which is mentioned Pair-wise Comparisons If the Kruskal-Wallis test rejects the null hypothesis of equality of treatment effects, it indicates that at least two of the treatment effects are unequal. It does not tell the researcher which one are different from each other. Therefore, a test procedure for making pair wise comparisons is needed. For this, the null hypothesis H 0 : τ i = τ i against H : τ i τ i i i =,,..., can be tested at α % leel of significance by using the inequality 68

N( N + ) Ri Ri' z p i,i',i i' =, + ri r i',..., where p = α/(-) and z p is the quantile of order -p under the standard normal distribution. From the aboe, we can say that the least significant difference between the treatments i and i is c ii' = z Therefore, if p N( N + ) ri R > + r i' i Ri' cii' then the difference between i and i' treatment effects is considered significant at α% leel of significance. The aboe procedure is illustrated with the help of following example. Example.: An experiment was conducted with animals to determine if the four different feeds hae the same distribution of Weight gains on experimental animals. The feeds, 3 and 4 were gien to 5 randomly selected animals and feed was gien to 6 randomly selected animals. The data obtained is presented in the following table. Feeds Weight gains (kg) 3.35 3.8 3.55 3.36 3.8 3.79 4. 4. 3.95 4.5 4.4 3 4 4.5 4.5 4.75 5 4 3.57 3.8 4.09 3.96 3.8 We use Kruskal-Wallis test to analyze the aboe data. We arrange the data in ascending order and gie the ranks to to the obserations. The ranks are then arranged feed wise as gien below: Feeds Ranks of Weight gains Sum of ranks (R i ) Aerage of Ranks 6 3 7 9 3.80 5 4 5 0 6 7 77.83 3 8 9 0 90 8.00 4 4 9 3 8 45 9.00 Then the Kruskal-Wallis test statistic is obtained as: T (9 ) = * 5 (77 ) + 6 ( 90 ) + 5 ( 45 ) + 5 3* = 80.4 66 = 4.4 The tabulated alue of χ at 3 degree of freedom at 5% leel of significance is 7.85 and the calculated alue is 4.4 so we infer that feed effects are differing significantly. 683

Pair-wise Comparisons for the feeds Here r = r 3 = r 4 = 5; r = 6, N =, = 4, and (-) =. Let α = 0.05 then p = 0.5/6 = 0.00466. For this alue of p, z p =.6. We can compute c ii as * c = c3 = c4 =.6 + = 9.77 5 6 * c3 = c4 = c34 =.6 + = 0. 5 5 Thus R R = 9.03; R R3 = 4.; R R4 = 5.; R R3 = 5.7; R R4 = 3.83; R3 R4 = 9. Here we see that R R3 > c 3 so the effect of feeds and 3 are significantly different at 5% of leel of significance while all other pairs of feeds effects do not differ significantly.. Friedman test The Kruskal-Wallis test is useful for the data generated through completely randomized designs. A completely randomized design is used when experimental units are homogeneous in a block. Howeer, there do occur experimental situations where one can find a factor (called nuisance factor), which, though not of interest to the experimenter, does contribute significantly to the ariability in the experimental material. Various leels of this factor are used for blocking. For the experimental situations where there is only one nuisance factor, the block designs are being used. The simplest and most commonly used block design by the agricultural research workers is a randomized complete block (RCB) design. The problem of non-normality of data may also occur in RCB design as well. Friedman test is useful for such situations (see Friedman, 937). Let there are treatments that are arranged in N = b experimental units arranged in b blocks of size each. Each treatment appears exactly once in each block. The data generated through a RCB design can analyzed by the following linear model y ij = µ + τ i + β j + ij, i =,,, ; j =,,, b. where y ij is the yield (response) of the i th experimental unit receiing the treatment in j th block. τ i is the effect due to i th treatment. β j is the effect of j th block. ij is random error in response. Now we are interested to test the equality of treatment effect. In other words, we want to test the null hypothesis H 0 : τ = τ = τ 3 = = τ =τ (say) against the alternatie H : at least two of the τ i s are different. For using Friedman test, we proceed as follows. Arrange the obserations in rows (treatments) and b columns (blocks). The obserations in the different rows are independent and those in different columns are dependent. Rank all the obserations in a column (block) i.e. ranks are assigned separately for each block. Let R ij be the rank of the obseration pertaining to i th treatment in the j th block. Then R ij. As the ranking has been done within blocks from to, therefore sum of ranks in j th block is 684

R j = R ij ( + ) = and treatment is R i = R ij. b j= + R j = and the ariance is. Sum of ranks for each If the treatment effects are all the same then we expect each R i to be equal b(+)/, that is, under H 0, b( + ) b( + ) Ε ( Ri ) = =. The sum of squared deiations of R i s from E(R i ) is, therefore, a measure of the differences in the treatment effects. Let S = b( + ) Ri The Friedman test statistic is then defined as T S b( + ) = = Ri b( + ) b( + ) = Ri 3b( + ) χ b( + ) ( ) The method of determining the probability of occurrence when H 0 is true of an obsered alues of T depends upon the sizes of and b. For small alues of b and, the null distribution of T has been tabulated. For a ready reference, the table has been included in Appendix-II. For large b and, the associated probability may be approximated by the χ distribution with - degrees of freedom. Remark.: When there are ties among the ranks for any gien block, the statistics T must be corrected to account for changes in the sampling distribution. So if ties occur then we use following statistic Ri 3b ( + ) T = χ (-) b g j 3 b t js j= s= b( + ) + ( ) 685

where g j is the number of sets of tied ranks in the j th block and t js is the size of the j th set of tied ranks in the i th block. Pair-wise Comparisons When the Friedman test rejects the null hypothesis that the all treatment effects are not the same, it is of interest to identify significant difference between the paired treatments. Therefore, a test procedure for making pair wise comparisons is needed. The null hypothesis H 0 : τ i = τ i against H : τ i τ i i i =,,..., can be tested at α % leel of significance using the inequality. b( + ) Ri Ri' z p for all i,i' =,,...,,i i' 6 where p = α/(-) and z p is the quantile of order -p under the standard normal distribution. From the aboe, we can say that the least significant difference between the treatments i and i is b( + ) c = z p 6 If R i Ri' > c then the difference between treatment i and i' are considered as significantly different at α% leel of significance. The aboe procedure is illustrated with the help of following example: Example.: An animal feeding experiment inoling 8 different rations was laid out in a RCB design using 4 animals in 3 groups of size 8 each. The grouping was done on the basis of initial body weight. Block Rations 3 38.46 400 46.5 387.6 369. 384.6 3 436.9 430.8 467.7 4 44.6 46.5 4.5 5 430.7 455.3 406. 6 44.6 384.6 384.5 7 44.59 350.8 307.6 8 360 400. 400 Now first we check the normality of obserations. For testing the normality of obseration we use the Shapiro-Wilk Test and Kolmogoro-Smirno test. Here H 0 : Obserations come from normal population against H : Obserations do not come from normal population. Kolmogoro-Smirno Shapiro-Wilk Statistic df Sig. Statistic df Sig. Residual.00 4.04.873 4.00 686

We can see that the obserations are non-normal at 5% leel of significance. Now we use usual method of ANOVA. We get Sum of Source DF Squares Mean Square F Value Pr > F Replication 3889.956 944.9780 0.40 0.6747 Treatment 7 3075.44 458.774 0.95 0.4988 Error 4 6773.404 4805.443 Total 3 0338.690 From this analysis we can see that treatments are significantly different at 0% leel of significance. Now we use the Friedman Test for analysis of same data. We rank the data within each block then we get Block Sum of ranks Rations 3 (R i ) 4 7 3 3 8 3 8 6 8 4 5 8 6 9 5 7 7 5 9 6 6 3 7 4 6 8 5 4 Here b = 3 and = 8, so the Friedman statistic is T = 3* 8( 8 + ) = 94-8 = 3 [( ) + ( 8 ) + ( ) + (9 ) + (9 ) + () + (6 ) + () ] 3* 3* 9 The tabulated alue of χ at 7 degree of Freedom is.0 and the calculated alue is 3. So the rations are significantly different at 0% leel of significance. Probability greater than χ is 0.07084. Pair-wise Comparisons Here b =3, = 8 and (-) = 56. Let α = 0. then p = 0./56 = 0.007857. z p =.. Then we calculate 3* 8( 8 + ) c =. =.6 6 687

So the critical difference of rank sum is.6, if R i R i' >.6 the treatment i and i' is significantly different. For example R R 3 = 4 is more than.6, so we conclude that it is significantly different at 0% leel of significance..3 Durbin s Test In the section., we hae discussed the non-parametric analysis of experimental data of a RCB design. In a RCB design, the number of experimental units required in each block are same as the number of treatments. Howeer, when the number of treatments increase, the blocks become large and it is not possible to maintain homogeneity with blocks. If an experimenter persists with a RCB design it results into large intra block ariances and hence reduced precision on treatment comparisons. To circument this problem, recourse is made incomplete block designs. Many a times an experimenter may hae to use incomplete block designs because of the nature of experimental units. The simplest of the incomplete block designs are balanced incomplete block (BIB) design. This standard notations for describing a BIB design is gien below: = the number of treatments b = the number of blocks r = number of replications of each treatment k = number of experimental units per block (k<) λ= the number of blocks in which a gien treatment pair occurs together The model of a BIB design is same as gien in section.. For a non-normal data situation pertaining to a BIB design, Durbin (95) proposed a test statistic. We want to test the equality of the treatment effects i. e. the null hypothesis H 0 : τ = τ = τ 3 = = τ =τ (say) against the alternatie H : at least two of the τ i s are different. To test the aboe hypothesis, rank the obserations y ij from to k within a block. Let R ij be the rank of y ij. Then following the lines of Friedman test, Durbin test statistic is gien by T = r( k ( ) r( k + ) Ri )( k + ) ( ) 3r( )( k + ) Ri ~ k )( k + ) ( k ) = χ r( ( ) The test rejects H 0 if T is more than the cut-off point. The cut off point is obtained by referring to the chi-square distribution with (-) degree of freedom. The exact test can be obtained by rejecting H 0 when T m α, where some alues of m α are gien in Skillings and Mack (98). Pair-wise Comparisons When the Durbin test rejects the null hypothesis that the all treatment effects are not the same, it is of interest to identify significantly different treatments. Therefore, a test procedure 688

for making pair wise comparisons is needed. The null hypothesis H : i τ i H 0 : i = τ i τ i i =,,..., can be tested at α % leel of significance using the r( k + )( k ) Ri Ri' z p i i',i,i' =,,..., 6( ) τ against where p = α/(-) and z p is the quantile of order -p under the standard normal distribution. From the aboe, we can say that the least significant difference between the treatments i and i is r( k + )( k ) c = z p 6( ) If R i Ri' > c then the difference between treatments i and i' is considered significant at α s leel of significance. The aboe procedure is illustrated with the help of following example: Example.3: In an experiment to compare palatability of four arieties of rice (cooked), four judges were asked to rank three arieties each. The results are as follows. Rice Judges Sum of Variety 3 4 ranks (R i ) - 4 3-6 3-5 4-3 3 3 9 Now we the use the Durbin test ( 4 ) T = 3* 4 * ( 3 )( 3 + ) [( 4 6) + ( 6 6) + ( 5 6) + ( 9 6) ] = 5.5 The tabulated alue of χ at 3 degree of Freedom is 7.85 and the calculated alue is 5.5 so the treatments are not significantly different at 5 % leel of significance. Here the probability greater than χ is 0.074398..4 Skillings and Mack Test In some of the experimental situations, the use of a BIB design may not be feasible. One may hae to use a partially balanced incomplete block (PBIB) design. In some of the experimental situations een a non-proper block design may be useful. Skillings and Mack (98) proposed a Friedman-type test statistic i. e. useful for any binary block design. Let (, b, r, k) represents a binary block design in which treatments arranged in b blocks such that j th block contains k j distinct treatments and i th treatment is replicated r i times; i =,,,, j =,,, b. For the analysis of experimental data generated through a binary block designs, 689

we make use of statistic gien by Skillings and Mack (98). To compute the test statistic, we find adjusted treatment sums for ranked data. For this we proceed as follows:. Within each block, rank the obserations from to k j, where k j is the number of treatment present in j th block.. Let R ij be the rank of y ij if the obseration is present; otherwise, let R ij = (k j +)/ 3. Compute an adjusted treatment sum for the i th treatment, namely A b = / [ /( k + ) ] [ R ( k ) / ] i j ij j + j= Let matrix, which is the coariance matrix of the random ector A' = (A,, A ). The coariance structure of the R ij s is well known and in this case only minor modifications are required because of missing cells. In block j, under H 0 : τ = τ = τ 3 = = τ = τ (say), we hae ( k j + )( k j ) / if treatment i is present in block j Var( Rij ) =, 0,otherwise ( k j + ) / If j = j', i i' nij = and ntj' = Co( Rij,Ri' j' ) =, 0,otherwise where n ij is the number of times treatment i appears in block j. Thus and Var( A b i ) = ( k j )nij, j= Co( A b i, Ai' ) = nij nii', j= i =,,, i i' where n ij equals one if treatment i appear in block j and equal to zero otherwise. By defining λ it to be the number of blocks containing obserations for both treatments i and i'. It can be seen by inspection that under H 0 the ((σ ii' )) can be rewritten as and σ ii' = -λ ii', i i' σ ii = i' = i' i λ ii' = i' = i' i σ ii', i =,,,. Thus the elements of are simple to obtain. The off-diagonal elements are (-λ ii' ), and the diagonal elements are the negatie of the sum of off-diagonal elements in that row. We note that the coariance matrix is singular, because the sum of the rows (columns) is always zero. In any connected block design, the rank of will be -. The test statistic we now propose is of the form T =A' - A 690

where - is a generalized inerse of. T follows an approximate χ -distribution with d.f. as the rank of. The aboe procedure is illustrated with the help of following example: Example.4: An engineer is studying the mileage performance characteristics of 4 types of gasoline addities. In the road test he wishes to use cars as block (9). The results are as follows. Gasoline Car (Block) Addities 3 4 5 6 7 8 9 A 3. 3. 4.3 3.5 3.6 4.5 4.3 3.5 B 4. 3.9 3.5 3.6 4. 4.7 4. 4.6 C 3.8 3.4 4.6 3.9 3.7 3.7 3.4 4.4 3.7 D 4. 4 4.8 4 3.9 4.9 3.9 Now we rank the data within each block. If obseration is not present then we use R ij = (k j +)/. We get Gasoline Car ( Block) Addities 3 4 5 6 7 8 9 A.5* B 3 3 4 3 3 * C 3 3 D 4 4 4 4 3 *.5* 4 3 * Represents missing obseration rank. Now we calculate adjusted treatment sums (A i ) as Block Adjusted 3 4 5 6 7 8 9 Treatment Sum (A i ) A -.338 -.338-0.7746 -.338 -.338 0.0000 0.0000 -.338 -.73-4.56 B 0.7746 0.7746 -.338-0.7746.338.73.0000 0.7746 0.0000 4.8 C -0.7746-0.7746 0.7746 0.7746-0.7746 -.73 -.0000-0.7746 0.0000-4.8 D.338.338.338.338 0.7746 0.0000 0.0000.338.73 4.56 So, A i = (-4.56 4.8-4.8 4.56)' The coariance matrix is obtained by counting the number of times treatment pair occur together. 69

7 7 = 8 8 7 6 And Generalized inerse - is 8 8 3 7 7 6 7 0 49 5 = 4 0 5 44 3 0 4 3 43 0 0 0 0 0 Now, T = A' - A = [-4.56 4.8-4.8 4.56] = 5.49 49 5 4 0 5 44 3 0 4 3 43 0 0 4.56 0 4.8 0 4.8 0 4.56 The tabulated alue of χ at 3 degree of Freedom is 7.85 and the calculated alue is 5.49. So the treatments are significantly different at 5 % leel of significance. Here the probability greater than χ is 0.00444. The statistic is quite general and the commonly used Friedman test statistic and Durbin test statistic discussed in Section. and.3 respectiely are the particular cases of this test. For example, in a RCB designs all n ij =, all λ ii' = b and k j =, therefore, for a RCB design = b(i - ') and A i = Substituting, these in T, we get T = ( b ) Ai T = Ri 3b( + ) b( + ) / b + Rij + j= Now in case of a BIB design, all blocks hae k< obserations and the number of blocks in which any pair of treatments occur together is, λ. Therefore, for a BIB design = (r(k- )+λ)i-λ '. T statistic is reduces to. 69

T T = ( λ ) Ai ( ) 3r( )( k + ) = Ri r( k )( k + ) k Since λ(-) = r(k-).5 Gore test for Multiple Obserations per Plot The test discussed aboe can also be used for general block design where only one obseration per plot is aailable. Howeer sometime more than one obserations are aailable from each cell. The appropriate model for such situation is y ijk = µ + τ i + β j + ijk, i =, ;, ; j =,,, b; k =,,, n ij. where y ijk is the k th obseration in the (i, j) th cell, n is the number of obserations in the (i, j) th cell, that is the number of experimental units receiing i th treatment and i th block. ijk are independent errors that follow a continuous distribution with a zero median. We are interested to test the equality of the treatment effect i.e. H 0 : τ = τ = τ 3 = = τ = τ (say) against an alternatie hypothesis that at least two of the τ i s are different. b Now, suppose N = nij, the total number of obserations and i j nij pij =, N pij q ij =, qi. = qij, b j * q.. = q i i. Consider a pair of plots (i, j) and (i, j). They are in the same column j. We can form n ij.n i j pairs of obserations by taking one obseration from each of these cells. Suppose u i,i,j is the proportion of these pairs, such that the obseration from plot (i,j) is larger of two. Then define b u i = ui'ij i' j= Using these, the test statistic is N * T = ( ui ( )b / ) / qi. ( ui ( )b / ) / q i. / q.. The test rejects H 0 if T is greater than the upper cut off point of the chi-square distribution with ( -) degree of freedom. The aboe procedure is illustrated with the help of following example: Example.5: The following table gies the number of days to maturity for three arieties of a cereal crop grown in two soil conditions. 693

Soil type Variety Light Heay A 30, 5, 3, 4 7, 5, 39 B 08, 4, 4, 06 9,, 0 C 55, 46, 5, 65 97, 08 In this data set n = 4, n = 3, n = 4, n = 3, n 3 = 4, n 3 =. and = 3, b =, N = 0, Now Then p = p = p 3 = 4/0 = 0.; p = p = 3/0 = 0.5; p 3 = /0 = 0. q = q = q 3 = /0.0 = 5; q = q = /0.5 = 6.67; q 3 = /0. =0 q. = 5+6.67=.67; q. = 5+6.67=.67; q 3. = 5+0=5 q.. = + + = 0.380.67.67 5 Computation of u i,i,j : Let us first compute u,,, here 6 (4x4) comparisons altogether. Take each obseration in cell (,) and compare it with 4 obserations in cell (, ). Obseration 30 is bigger than all 4 alues in cell (, ). Hence, it contributes 4. Similarly 5 contribute 3, 3 contributes 3 and 4 contributes 4. The total is 4. So, u,, is 4/6. Similarly u,, =9/9; u,3, =0; u,3, =; u,3, =0; u,3, =4/6; u,, =/6; u,, =0; u 3,, =; u 3,, =0; u 3,, =; u 3,, =/6; Hence 4 9 46 4 76 4 u = + + 0 + = ; u = + 0 + 0 + = ; u 3 = + 0 + + = 6 9 6 6 6 96 6 6 From these now we calculate test statistic N * T = ( ui ( )b / ) / qi. ( ui ( )b / ) / q i. / q.. * 0 T = [( 0.0656 + 0.5 + 0.0074) ( 0.0656 + 0.5 + 0.0074) / 038] 3 = 5.8 The tabulated alue of χ at degree of Freedom is 5.99 and the calculated alue is 5.8 so the arieties of cereal crop are not significantly different at 5% leel of significance. Here the probability greater than χ is 0.0736. A lot of efforts hae also been made at IASRI to deelop the non-parametric test procedures for the analysis of groups of experiments conducted in Randomized block designs and split plot designs. For details one may refer to Rai and Rao (980, 984). Acknowledgements: The help receied from Sh. Ajeet Kumar, M.Sc. (Agricultural Statistics) student during the preparation of this lecture note is duly acknowledged. 694

References Conoer, W.J. and Iman, R.L. (976). In Some Alternatie Procedure Using Ranks for the Analysis of Experimental Designs. Commun. Statist. Theor. Math. A5(4), 349-368. Deshpande, J.V., Gore, A.P. and Shanubhogue, A. (995). Statistical Analysis of Nonnormal Data. Wiley Eastern Limited, New Delhi. Durbin, J. (95), Incomplete Blocks in Ranking Experiments. Brit. J. Statist. Psych. 4, 84-90. Friedman, M. (937), The Use of ranks to aoid the assumption of Normality implicit in the Analysis of Variance, J. Amer. Statist. Assoc. 3, 675-70. Gore, A.P. (975). Some Non-parametric Tests and Selection Procedures for Main Effects in Two- ways Layouts. Annals of the Institute of Statistical Mathematics, 7, 487-500. Gore, A.P. and Shanobhogue, A. (988). A Note on Rank Analysis of Split Plot Experiments, Journal of Indian Society of Agricultural Statistics, XL 3, 78-84. Hollander, M. and Wolfe, D.A. (999). Nonparametric Statistical Methods, nd Ed, New York: John Wiley. Kruskal, W.H. (95). A Nonparametric Test for the Seeral Sample Problem, Ann. Math. Statist. 3, 55-540. Kumar, Ajeet. (00). Non-parametric methods in analysis of experimental data. Course Seminar Write deliered during January April, 00. Rai, S.C. and Rao, P.P (980). Use of Ranks in Groups of Experiments. Indian Soc. Agril. Statist., 3, 5-3. Rai, S.C. and Rao, P.P (984). Rank Analysis of Group off Split-Plot Experiments. Indian Soc. Agril. Statist., 36, 05-3. Sen, P.K. (996). Design and Analysis of Experiments: Non-parametric Methods with Applications to Clinical Trials. Handbook of Statistics, Vol 3, eds. S.Ghosh and C.R.Rao, 9-50. Siegel, S. and Catellan, N.J. Jr. (988). Non parametric statistics for the Behaioural Sciences, nd Ed., McGraw Hill, New York. Skillings, J.H. and Mack G.A. (98). On the Use of a Friedman-Type Statistic in Baanced and Unbalanced Block Designs. Technometrics 3 (), 7-77. 695

Appendix-I Kruskal-Wallis Test Statistic Each table entry is the smallest of the Kruskal-Wallis T such that its right-tail probability is less than or equal to the alue gien on the top row for = 3, each sample size less than or equal to fie. r r r 3 0.00 0.050 0.00 0.00 0.00 4.57 - - - - 3 4.86 - - - - 3 4.500 4.74 - - - 3 3 4.57 5.43 - - - 3 3 4.556 5.36 6.50 - - 3 3 3 4.6 5.600 6.489 7.00-4 4.500 - - - - 4 4.458 5.333 6.000 - - 4 3 4.056 5.08 - - - 4 3 4.5 5.444 6.44 6.444-4 3 3 4.709 5.79 6.564 6.745-4 4 4.67 4.967 6.667 6.667-4 4 4.555 5.455 6.600 7.036-4 4 3 4.545 5.598 6.7 7.44 8.909 4 4 4 4.654 5.69 6.96 7.654 9.69 5 4.00 5.000 - - - 5 4.373 5.60 6.000 6.533-5 3 4.08 4.960 6.044 - - 5 3 4.65 5.50 6.4 6.909-5 3 3 4.533 5.648 6.533 7.079 8.77 5 4 3.987 4.985 6.43 6.955-5 4 4.54 5.73 6.505 7.05 8.59 5 4 3 4.549 5.656 6.676 7.445 8.795 5 4 4 4.668 5.657 6.953 7.760 9.68 5 5 4.09 5.7 6.45 7.309-5 5 4.63 5.338 6.446 7.338 8.938 5 5 3 4.545 5.705 6.866 7.578 9.84 5 5 4 4.53 5.666 7.000 7.83 9.606 5 5 5 4.560 5.780 7.0 8.000 9.90 696

Appendix-II Friedman Test Statistic Critical alue for the Friedman two-way analysis of ariance by rank statistics, T b α.0 α.05 α.0 3 3 4 5 6 7 8 9 0 3 6.00 6.00 5.0 5.33 5.43 5.5 5.56 5.00 4.9 5.7 4.77 6.00 6.50 6.40 7.00 7.4 6.5 6. 6.0 6.54 6.7 6.00 --- 8.00 8.40 9.00 8.86 9.00 8.67 9.60 8.9 8.67 9.39 4.6 5.99 9. 4 3 4 5 6 7 8 6.00 6.00 --- 6.60 7.40 8.60 6.30 7.80 9.60 6.36 7.80 9.96 6.40 7.60 0.00 6.6 7.80 0.37 6.30 7.50 0.35 6.5 7.8.34 5 3 4 5 7.47 8.53 0.3 7.60 8.80.00 7.68 8.96.5 7.78 9.49 3.8 697