Downloaded from https://www.everyspec.com AMCP 706.198 Example No. 10 (Cont'd) 4. Calculate the values of c which form 4. See Table 2-11 for results. the boundaries c= 16. 6ke ~-. (the inve rse of Eq. 2-38). From the data find how many are In each cell. The n,3 Is 5, because the (ASP x 50 - 3, 5. Find X 2 , v. Use Eq. 2--44 for X 2 5. X 2 9.20 for the sample (same answer k v L- k - s- 1. 2 param= as in Table 2-11). eters (a and g) were determined. 2, ke 10 9 7. 6. Find csqf(X 2 , 1 :) from Table 2-11. 6. csqf(9.20; 0) P 0.58 Numbers very near one cause the null hypothesis to be rejected.
csqf(9.20;7) 0,(5 Thus, values of X 2 would, by chance alone, be bigger than 9.2 about 25% to 40% of the time. 7. Is the result s-significant; i.e. , 7. Our value of 1), 2 is reasouable anud we how unlucky were we? do not reject the null hypothesis that "the lives in Trble 2-1.0 are fronm the Weibull distribultion calculated in step 1." The results are not s-stIrliflttlt even at the 20% level. ...........