41. The single-factor ANOVA model considered in Chapter 11 assumed the observations in the ith sample were selected from a normal distribution with mean mi and variance s2, that is, Xij  mi  eij where the e s are normal with mean 0 and variance s2. The normality assumption implies that the F test is not distribution-free. We now assume that the e s all come from the same continuous, but not necessarily normal, distribution, and develop a distribution-free test of the null hypothesis that all I mi s are identical. Let , the total number of observations in the data set (there are Ji observations in the ith sample). Rank these N observations from 1 (the smallest) to N, and let be the average of the ranks for the observations in the ith sample. When H0 is true, we expect the rank of any particular observation and therefore also to be (N  1)/2. The data argues against H0 when some of the s differ considerably from (N  1)/2. The Kruskal–Wallis test statistic is When H0 is true and either (1) I  3, all Ji  6 or

(2) I 3, all Ji  5, the test statistic has approximately a chi-squared distribution with I  1 df.

The accompanying observations on axial stiffness index resulted from a study of metal-plate connected trusses in which ve different plate lengths 4 in., 6 in., 8 in., 10 in., and 12 in. were used ( Modeling Joints Made with Light-Gauge K 

12 N1N  12 aJi aRi 

N  1 2

b 2

Metal Connector Plates, Forest Products J., 1979:
39—44).
i  1 (4): 309.2 309.7 311.0 316.8 326.5 349.8 409.5 i  2 (6): 331.0 347.2 348.9 361.0 381.7 402.1 404.5 i  3 (8): 351.0 357.1 366.2 367.3 382.0 392.4 409.9 i  4 (10): 346.7 362.6 384.2 410.6 433.1 452.9 461.4 i  5 (12): 407.4 410.7 419.9 441.2 441.8 465.8 473.4 Use the K—W test to decide at signi cance level .01 whether the true average axial stiffness index depends somehow on plate length.