Fundamentals of Probability and Statistics for Engineers

322 Fundamentals of Probability and Statistics for EngineersTable 10.3 Table for 2 test for Example 10.2Interval, A i n i p i np i n 2 i /np ix 0 5147 0.6188 4853 54590 < x 1 1859 0.2970 2329 14841 < x 2 595 0.0713 559 6332 < x 3 167 0.0114 89 3133 < x 4 54 0.0013 10 2924 < x 5 14 0.0001 1 1965 < x 6 0.0001 1 367842 1.0 7842 8413Note: n i , observed number of occurrences; p i ,theoretical P(A i ).We thus haveP…A i †ˆp i ˆ …0:48†i 1 e 0:48; i ˆ 1; 2; ...; 6;…i 1†!P…A 7 †ˆp 7 ˆ 1X 6iˆ1p i ;These values are indicated in the third column of Table 10.3.Column 5 of Table 10.3 givesd ˆ Xkiˆ1n 2 inp in ˆ 8413 7842 ˆ 571:2 2With k ˆ 7, the value of k 1,ˆ is found from Table A.5 to be 6, 0:0126 0 01 ; : ˆ 16: 812:2Since d > 6, 01 , the hypothesis is rejected at the 1% significance level.0:10.2.2 THE CASE OF ESTIMATED PARAMETERSLet us now consider a more common situation in which parameters in thehypothesized distribution also need to be estimated from the data.A natural procedure for a goodness-of-fit test in this case is first to estimatethe parameters by using one of the methods developed in Chapter 9 and then tofollow the 2 test for known parameters, already discussed in Section 7.2.1. InTLFeBOOK