When data are nonnormal in form classical procedures for assessing treatment group equality are prone to

distortions in rates of Type I error and power to detect effects

Fradette, Katherine

Keselman, H. J.

Othman, Prof.Madya Abdul Rahman

Padmanabhan, A. R.

Wilcox, Rand R.

English

Repository@USM

Journal of Modern Applied Statistical Methods  Copyright  2002 JMASM, Inc. Fall 2002, Vol. 1, No 2, 310-315  1538 – 9472/02/$30.00 310 A Test Of Symmetry  Abdul R. Othman Universiti Sains Malaysia  H. J. Keselman Dept. of Psychology University of Manitoba  Rand R. Wilcox Dept. of Psychology Univ. of Southern California  Katherine Fradette University of Manitoba  A. R. Padmanabhan Monash University, Austrailia    When data are nonnormal in form classical procedures for assessing treatment group equality are prone to distortions in rates of Type I error and power to detect effects. Replacing the usual means with trimmed means reduces rates of Type I error and increases sensitivity to detect effects. If data are skewed, say to the right, then it has been postulated that asymmetric trimming, to the right, should be better at controlling rates of Type I error and power to detect effects than symmetric trimming from both tails of the data distribution. Keselman, Wilcox, Othman and Fradette (2002) found that Babu, Padmanabhan and Puri's (1999) test for symmetry when combined with a heteroscedastic statistic which compared either symmetrically or asymmetrically determined means provided excellent Type I error control even when data were extremely heterogeneous and very nonnormal in form. In this paper, we present a detailed discussion of the Babu et al. procedure as well as a numerical example demonstrating its use.  Key words: Symmetry, Preliminary test   Introduction  Keselman, Wilcox, Othman and Fradette (2002) found that by utilizing a test for symmetry prior to testing for equality of trimmed means they were able to achieve excellent Type I error control even though data were extremely heterogeneous and very nonnormal in form. In particular, they used a test  for  symmetry first proposed by Hogg, Fisher,     Abdul Rahman Othman is a lecturer in the School of Distance Education. His areas of interests are psychometrics and applied statistics. H. J. Keselman is Professor of Psychology. Email: kesel@ms.umanitoba.ca. Rand R. Wilcox is Professor of Psychology. Email: rwilcox@usc.edu. Katherine Fradette is an undergraduate honors student in the Department of Psychology. Her plans include graduate work in quantitative methods. A. R. Padmanabhan teaches in the Department of Mathematics in Australia. Work on this project was supported by the Natural Sciences and Engineering Council of Canada.  and Randles (1975) and subsequently modified by Babu, Padmanaban and Puri (1999) in order to determine whether data should be trimmed symmetrically or asymmetrically. Asymmetric trimming has been theorized to be potentially advantageous when the distributions are known to be skewed, a situation likely to be realized with behavioral science data (See De Wet & van Wyk, 1979; Micceri, 1989; Tiku, 1980, 1982; Wilcox, 1995). That is, theoretical considerations suggest that when data are say skewed to the right then in order to achieve robustness to nonnormality and greater sensitivity to detect effects one should trim data just from the upper tail of the data distribution. Indeed, Keselman et al. found that by combining a test for mean equality with a preliminary test for symmetry Type I error rates could be substantially improved for the nonnormal and heterogeneous distributions they examined. Because space considerations prevented them from providing a full description of the symmetry test we present the method herein and illustrate its application with a numerical example.   OTHMAN, KESELMAN, WILCOX, FRADETTE, & PADMANABHAN   311Theoretical Background  The Babu et al. (1999) procedure is based, in part, on the work of Hogg et al. (1975). Specifically, for these authors, the hypothesis of interest was H0:  = 0 against HA:   > 0, where   is the location parameter of interest. They proposed a test to detect the nature of the underlying distribution before proceeding with (nonparametric) tests of H0.  In particular, they defined Y1, Y2,…, Ym as a random sample from F(y), and Ym+1, Ym+2,…, Yn as a random sample from F(y - ). Then Y(1), Y(2),…, Y(n) are the ordered statistics of the combined random samples and Ymed is the median of the combined samples.   Hogg et al.’s (1975) procedure to detect the nature of the underlying distribution is composed of two tests, a test of the heaviness of the tail of the distribution using the Q2 statistic and a test of symmetry using the Q1 statistic. Their work was based on papers by Uthoff (1970, 1973). Hogg et al. (1975) chose a test statistic enumerated by Uthoff (1973, Equation 2) as a basis to define their Q2 index. This index determined whether the tail of the underlying distribution is light or heavy. They first approximated it as  (n) (1)(i) medY   - Y  .2 Y -Y / n  They transformed this ratio into   0.05 0.0520.5 0.5(U L )Q ,(U L )  where U0.05 and L0.05 are, respectively, the means of the upper and lower 5% of the order statistics of the sample and U0.5 and L0.5 are, respectively, the means of the upper and lower 50% of the order statistics of the combined sample.  Again, based on the work of Uthoff (1970, Equation 1), Hogg et al. (1975) derived their Q1 index:   0.0510.05( ),( )U MIDQMID L  where MID is the mean of the middle 50% of the combined sample. Thus, this index determines the symmetry of the underlying distribution.  Babu et al. (1999) extended the use of these two indices to more than two groups. They proposed that both indices be calculated within the groups and weighted means of these indices be the overall estimates of Q2 and Q1. They also proposed adjustments to the Q1 index whereby the amount of data needed to calculate the index depended on the outcome of the calculation of the Q2 index.   Determination of Symmetry  Consider the problem of comparing distributions F1 = F2 = … = FJ. One way of approaching this problem is to consider the one-way ANOVA problem of comparing means μ1 = μ2  = … = μJ from J distributions F1(y) = F(y- μ1), F2(y) = F(y- μ2), … , FJ(y) = F(y- μJ). When the distributions are unknown and one cannot assume that they are normal with equal variances, Babu et al. (1999) suggested the following procedure to determine heavy-tailedness and symmetry prior to applying the appropriate test on the location parameters:  Let jij 1 j 2 j n jY (Y ,Y ,...,Y )  be a sample from an unknown distribution Fj. Let j(1) j (2) j (n ) jY Y Y    represent the ordered observations associated with the jth group. Let   be the proportion of the data in the sample that are of interest as either the proportion of data to be trimmed or the proportion of data to be used in the calculation of several intermediate variables leading to two statistics, namely Q2 and Q1. Let g = [nj] +1, where [x] represents the greatest integer less than nj and r = g - nj. It is important to note that trimming here, and the amount trimmed, is just for purposes of assessing symmetry.  Q2 Index Prior to determining the symmetry of the distributions, the nature of their tails is examined. The Q2 index determines whether F1(y), F2(y),…, FJ(y) are normal-tailed, heavy-tailed or very heavy-tailed. Tail classification is determined in the following manner: 1. Define Uj and Lj as the means of the upper and lower nj order statistics, respectively, of the sample Yj. A TEST OF SYMMETRY  312Case 1. If nj   1,  then Uj = Yj(n ) j and Lj = Y (1) j .  Case 2. If nj  > 1  then   jjjn, j (i) j (n g 1), jj n g 2j1U Y (1 r)Yn  	  	  	        and  g 1, j (i) j (g) ji 1j1L Y (1 r)Yn  	        2. Calculate U0.05, j and L0.05, j as the mean of the upper and lower 0.05nj order statistics of Yj, respectively. 3. Calculate U0.5, j and L0.5, j as the mean of the upper and lower 0.5nj order statistics of Yj, respectively. 4. For each j, set Q2, j  = (U0.05, j  -  L0.05, j ) / (U0.5, j  -  L0.5, j ).  5. Using Q2, j, j = 1, 2,…, J, from # 4 compute  J J2 j 2, j jj 1 j 1Q n Q n               6. If Q2 <  3 then F is classified as normal-tailed. If 3    Q2 <  5 then F is classified as heavy-tailed. If Q2   5 then F is classified as very heavy-tailed.  Q1 Index  Once the nature of the tails of the distributions is known, the Q1 index, which determines the symmetry of the distributions, is calculated. To calculate the Q1 index one should: 1. Based on Q2, determine the number of sample points in each sample Yj to be used. Define this as nj*. (This is the Babu et al., 1999, modification of the Hogg et al., 1975, proposal for computing Q1.) Specifically, if Q2 <  3 then use all sample points in Yj. If 3    Q2 <  5  then trim the top and bottom 10% of the sample points and use the middle 80% in Yj. If Q2   5  then trim the top and bottom 20% of the sample points and use the middle 60% in Yj. 2. Let MIDj to be the mean of the middle 50% of the order statistics of the sample points in sample Yj defined in #1. According to A. R. Padmanaban (personal communication, June 26, 2001), MIDj is calculated in the following manner:   Discard the top and bottom 25% of the order statistics of Yj.   The remainder is the middle 50% of the order statistics of Yj.  Hence, * *jg [ 0.25n ] 1 	  and * * *jr g 0.25n  . Therefore, MIDj is given by  * *j* * *j*n g*j (i) j* (g ) j (n g 1) ji g 1j1MID Y r (Y Y ) .0.5n 	 	  	 	     3.  For each j, set              1, j 0.05, j j j 0.05, jQ (U MID ) ( MID L )   .  4. Using Q1, j, j = 1, 2,…, J, from # 3 compute         jJ J* *1 j 1, jj 1 j 1Q n Q n               .  5. If Q1 < ½, F is deemed to be left skewed. If  ½  Q1  2, then F is considered to be symmetric. If Q1 > 2, then F is designated as right skewed.  Computational Example  Suppose we want to test the null hypothesis,  Ho: F1(x) = F2(x) = F3(x) based on the following data set.  Table 1. Data set.  Groups Order Statistics nj 1 30 32 32 34 35 35 39 40 40 41 42 48 50 52 99 152 35 36 40 40 41 42 43 49 56 64 103 48 48 51 51 51 55 55 60 63 83 10Note: The tabled values were chosen so that the data would be classified as heavy-tailed.    Calculating Q2 (Tail thickness)   Notice that 0.05nj < 1 for j = 1, 2, 3. Therefore, U0.05, 1  = Y(15, 1) = 99, U0.05, 2  = Y(10, 2)  = 64, U0.05, 3  = Y(10, 3) = 83, and L0.05, 1  = Y(1, 1) = 30, OTHMAN, KESELMAN, WILCOX, FRADETTE, & PADMANABHAN   313L0.05, 2  = Y(1, 2) = 35, and L0.05, 3  = Y(1, 3) = 48. When  = 0.5, the calculations for U0.5, j, L0.5, j and Q2, j for each group are as follows:  Group 1 n1 = 15, 0.5 n1 = 7.5, g = 8 and r = 0.5.  150.5,1 (i1) (8,1)i 91U Y 0.5Y7.51((40 41 99) (0.5)40)7.552.2667  	   	 	 	 	   70.5,1 (i1) (8,1)i 11L Y 0.5Y7.51((30 32 39) (0.5)40)7.534.2667  	   	 	 	 	   2,1(99 30)Q 3.8333(52.2667 34.2667)   Group 2  n2 = 10, 0.5 n2 = 5, g = 6 and r = 0.  100.5,2 (i2) (5,2)i 61U Y (0)Y51((42 43 64) 0)550.8  	   	 	 	 	   50.5,2 (i2) (6,2)i 11L Y (0)Y51((35 36 41) 0)538.4  	   	 	 	 	   2,2(64 35)Q 2.3387(50.8 38.4)    Group 3  n3 = 10, 0.5 n3 = 5, g = 6 and r = 0.  100.5,3 (i3) (5,3)i 61U Y (0)Y51((55 55 83) 0)563.2  	   	 	 	 	   50.5,3 (i3) (6,3)i 11L Y (0)Y51((48 48 51) 0)549.8  	   	 	 	 	   2,3(83 48)Q 2.6119(63.2 49.8)   Therefore,   2(15( 3.8333 ) 10( 2.3387 ) 10( 2.6119 ))Q(15 10 10 )3.0573	 		 	  and F is classified as heavy-tailed.  Calculating Q1   Because F if classified as heavy-tailed, we have to symmetrically trim 10% of the data before calculating Q1.  Notice that 0.05*jn < 1 for j = 1, 2, 3. Therefore:             * *0.05,1 (13,1)U Y 52  ,* *0.05,2 (8,2)U Y 56  ,            * *0.05,3 (8,3)U Y 63  , and * *0.05,1 (1,1)L Y 32  ,                 * *0.05,2 (1,2)L Y 36  , * *0.05,3 (1,3)L Y 48  .  Let us calculate MIDj and Q1, j, for j = 1, 2, 3.    A TEST OF SYMMETRY  314 Table 2. 10% Trimming.  Groups Order Statistics Following 10% Symmetric Trimming *jn1 32 32 34 35 35 39 40 40 41 42 48 50 52 13 2 36 40 40 41 42 43 49 56 8 3 48 51 51 51 55 55 60 63 8   Group 1  * * * *1 1n 13, 0.25n 3.25, g 4,and r 0.75    .  9* * *1 (i1) (4,1) (10,1)i 51MID Y 0.75(Y Y )6.51((35 39 40 40 41) (0.75)(35 42))6.538.8846  	 	   	 	 	 	 	 	  1,1(52 38.8846)Q 1.905(38.8846 32)   Group 2  * * * *2 2n 8, 0.25n 2, g 3,and r 0      6*2 (i2)i 31MID Y41(40 41 42 43)441.5     	 	 	  1,2(56 41.5)Q 2.6364(41.5 36)   Group 3 * * * *3 3n 8, 0.25n 2, g 3,and r 0    .  6*3 (i3)i 31MID Y41(51 51 55 55)453     	 	 	  1,3(63 53)Q 2(53 48)   Therefore,  1( 13(1.905 ) 8( 2.6364 ) 8( 2 ))Q(13 8 8 )2.133	 		 	   and F is classified as right skewed.   Discussion  As indicated in our introduction, Keselman et al. (2002) found that by first applying the Babu et al. (1999) procedure prior to testing for treatment group equality with sample symmetrically or asymmetrically determined trimmed means one could achieve excellent control over Type I errors even though data were obtained from very heterogenous distributions that were extremely nonnormal in form. Accordingly, they recommended that users adopt the Babu et al. (1999) test for symmetry.  It is also interesting to note that Babu et al. (1999) used the preliminary test for symmetry in order to determine whether groups should be compared on their symmmetrically determined trimmed means, when distributions were deemed symmetric, or on their medians, when distributions were deemed asymmetric. Thus, a test for symmetry can be beneficial in many different applications.  OTHMAN, KESELMAN, WILCOX, FRADETTE, & PADMANABHAN   315References  Babu, J. G., Padmanabhan A. R., & Puri, M. P. (1999). Robust one-way ANOVA under possibly non-regular conditions. Biometrical Journal, 413, 321-339. De Wet, T., & van Wyk, J. W. J. (1979). Efficiency and robustness of Hogg's adaptive trimmed means. Communications in Statistics, Theory and Methods, A8(2), 117-128. Hogg, R. V., Fisher, D. M., & Randles, R. H. (1975). A two-sample adaptive distribution free test. Journal of the American Statistical Association, 70, 656-661. Keselman, H. J., Wilcox, R. R., Othman, A. R., & Fradette, K. (2002). Trimming, transforming statistics, and bootstrapping: Circumventing the biasing effects of heterescedasticity and nonnormality. Journal of Modern Applied Statistical Methods, 1(2), 288-309). Micceri, T. (1989). The unicorn, the normal curve, and other improbable creatures.  Psychological Bulletin, 105, 156-166.                            Tiku, M. L. (1980). Robustness of MML estimators based on censored samples and robust test statistics. Journal of Statistical Planning and Inference, 4, 123-143. Tiku, M. L. (1982). Robust statistics for testing equality of means and variances. Communications in Statistics, Theory and Methods, 11(22), 2543-2558. Uthoff, V. A. (1970). An optimum test property of two well-known statistics. Journal of the American Statistical Association, 65, 1597-1600. Uthoff, V. A. (1973). The most powerful scale and location invariant test of the normal versus the double exponential. Annals of Statistics, 1, 170-174. Wilcox, R. R. (1995). ANOVA: A paradigm for low power and misleading measures of effect size? Review of Educational Research, 65(1), 51-77.    

A Test of Symmetry.

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A Test of Symmetry.

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