3.3.1 GENERAL TEST PROCEDURES 57 8. [OO] The (Web hosting reseller)
3.3.1 GENERAL TEST PROCEDURES 57 8. [OO] The text describes an experiment in which 20 values of the statistic K& were obtained in the study of a random sequence. These values were plotted, to obtain Fig. 4, and a KS statistic was computed from the resulting graph. Why were the table entries for n = 20 used to study the resulting statistic, instead of the table entries for n = 101 b 9. [%I] The experiment described in the text consisted of plotting 20 values of K&, computed from the maximum of 5 test applied to different parts of a random sequence. We could have computed also the corresponding 20 values of K,; since Kc has the same distribution as K&, we could lump together the 40 values thus obtained (that is, 20 of the K$ s and 20 of the K, s), and a KS test could be applied so that we would get new values K&, K,. Di scuss the merits of this idea. b 10. [XI] Suppose a chi-square test is done by making n observations, and the value V is obtained. Now we repeat the test on these same n observations over again (getting, of course, the same results), and we put together the data from both tests, regarding it as a single chi-square test with 2n observations. (This procedure violates the text s stipulation that all of the observations must be independent of one another.) How is the second value of V related to the first one? 11. [IO] Solve exercise 10 substituting the KS test for the chi-square test. 12. [A&?81 Suppose a chi-square test is made on a set of n observations, assuming that ps is the probability that each observation falls into category s; but suppose that in actual fact the observations have probability qs # ps of falling into category s. (Cf. exercise 3.) We would, of course, like the chi-square test to detect the fact that the ps assumption was incorrect. Show that this will happen, if n is large enough. Prove also the analogous result for the KS test. 13. [M.Z4] Prove that Eqs. (13) are equivalent to Eqs. (11). b 14. [HA&Z6] Let 2, be given by Eq. (18). Show directly by using Stirling s approxima- tion that the multinomial probability n!pyL..p;ti/Y,!…Y,! = e -V 2/~(2nn)k-1pl .pk + o(n-k 2), if Zr,Zs, . . . , zk are bounded as n + 00. (This idea leads to a proof of the chi-square test that is much closer to first principles, and requires less handwaving, than the derivation in the text.) 15. [HM24] Polar coordinates in two dimensions are conventionally defined by the equations x = r cos 8 and y = r sin 8. For the purposes of integration, we have dx dy = r dr do. More generally, in n-dimensional space we can let xk = r sin81 . . . sin&l cos&, l