use the Kruskal - Wallis test?. Car Crash Measurements? Use the following listed chest deceleration measurements (in g, where g is the force of gravity) from samples of small, midsize, and large cars. (These values are from Data Set 13 in Appendix B.) Use a 0.05 significance level to test the claim that the different size categories have the same median chest deceleration in the standard crash test. Do the data suggest that larger cars are safer? Small??44?39?37?54?39?44?42 Midsize?36?53?43?42?52?49?41 Large??32?45?41?38?37?38?33

Solution 7BSC Step 1 Kruskal-Wallis test satisfies the following two requirements. 1. Sample data must be independent and drawn according to simple random sampling. 2. Each sample must contain minimum of five entries. By using the data from exercise, we see that the samples are drawn at random, where the three samples are independent to each other. Hence the first requirement is satisfied. Now, we see that the sample size of Small is 7, sample size of Midsize is 7 and sample size of Large is 7. Hence the second requirement is satisfied. Therefore, the two requirements of Kruskal-Wallis test are satisfied. Step 2 By using = 0.05 level of significance we need t 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 different size categories have the same median chest deceleration in the standard crash test H1 the different size categories have the different median chest deceleration in the standard crash test Small44393754394442 Midsize36534342524941 Large32454138373833 Small Midsize Large 44(15.5) 36(3) 32(1) 39(8.5) 53(20) 45(17) 37(4.5) 43(14) 41(10.5) 54(21) 42(12.5) 38(6.5) 39(8.5) 52(19) 37(4.5) 44(15.5) 49(18) 38(6.5) 42(12.5) 41(10.5) 33(2) 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. The smallest value is assigned as the rank 1, here 32 is the smallest value and so on up to the largest value 54 is assigned as rank 21.