Correcting the H Test Statistic for Ties In using the Kruskal-Wallis test, there is a correction factor that should be applied whenever there are many ties: Divide H by For each individual group of tied observations in the combined set of all sample data, calcu?late T = ?t3? ? ?t?, where ?t? is the number of observations that are tied within the individual group. Find ?t? for each group of tied values, then compute the value of ?T? for each group, then add the Tvalues to get ? ?T?. The value of ?N? is the total number of observations in all samples combined. Use this procedure to find the corrected value of ?H? for Exercise. Does the cor?rected value of ?H? differ substantially from the value found in Exercise ? Exercise Nicotine in Cigarettes Refer to Data Set in Appendix B and use the amounts of nicotine (mg per cigarette) in the king-size cigarettes, the 100-mm menthol cigarettes, and the 100-mm nonmenthol cigarettes. The king-size cigarettes are nonfiltered, nonmenthol, and nonlight. The 100-mm menthol cigarettes are filtered and nonlight. The 100-mm nonmen?thol cigarettes are filtered and nonlight. Use a 0.05 significance level to test the claim that the three categories of cigarettes yield the same median amount of nicotine. Given that only the king-size cigarettes are not filtered, do the filters appear to make a difference? Mile 1 3:15 3:24 3:23 3:22 3:21 Mile 2 3:19 3:22 3:21 3:17 3:1 Mile 3 3:34 3:31 3:29 3:31 3:29

Solution 13BB Step 1 From Exercise 11, By using = 0.05 level of significance we need o test the claim that the different size categories have the same median chest deceleration in the standard crash test The Hypotheses can be expressed as H0 the three categories of cigarettes yield the same median amount of nicotine. H : the three categories of cigarettes yield the different median amount of nicotine. 1 All these three samples are combined together to get a big sample. We first order the absolute values of the difference scores and assign rank from 1 through ‘n’ to the smallest through largest absolute values of the difference scores, and assign the mean rank when there are ties in the absolute values of the difference scores. KgNic Rank MnNic Rank FLNic Rank 1.1 48 1.1 48 0.4 3 1.7 73.5 0.8 17 1 33.5 1.7 73.5 1 33.5 1.2 61 1.1 48 0.9 28 0.8 17 1.1 48 0.8 17 0.8 17 1.4 69 0.8 17 1 33.5 1.1 48 0.8 17 1.1 48 1.4 69 0.8 17 1.1 48 1 33.5 0.9 28 1.1 48 1.2 61 0.8 17 0.8 17 1.1 48 0.8 17 0.8 17 1.1 48 1.2 61 0.8 17 1.1 48 0.8 17 0.8 17 1.1 48 0.8 17 1 33.5 1.1 48 1.3 65.5 0.2 1.5 1.8 75 0.7 6.5 1.1 48 1.6 72 1.4 69 1 33.5 1.1 48 0.2 1.5 0.8 17 1.2 61 0.8 17 1 33.5 1.5 71 1 33.5 0.9 28 1.3 65.5 0.8 17 1.1 48 1.1 48 0.8 17 1.1 48 1.3 65.5 1.2 61 0.6 4.5 1.1 48 0.6 4.5 1.3 65.5 1.1 48 0.7 6.5 1.1 48 The smallest value is assigned as the rank 1, and so on up to the largest value is assigned as rank 75. The tied observation of the combined sample is as shown below Ties, Observation T = t t t 0.2 2 6 0.6 2 6 0.7 2 6 0.8 19 6840 0.9 3 24 1 8 504 1.1 21 9240 1.2 5 120 1.3 4 60 1.4 3 24 1.7 2 6 16,83 T 6 The Test Statistic can be expressed as 2 2 2 H = 12 [ R1 + R2 + . . . +k] 3(N + 1) N(N + 1) n1 n2 nk 2 650.5 786 H = 12 [ 1413.5 + + ] 3(75 + 1) 75(76) 25 25 25 H = 27.9098