Downloaded from https://www.everyspec.com AMCP 741-198 Snmimnov test Is to find the critical value of The procedure Is illustrated in Fig. 2-12(A) d from table 2-12 first. Then add It to and where the sample data from Table 2-13 are subtract it from the sample Cdf (the FRI and plotted. The line S is the result of plotting F&) In step 2b). This gives a band within FPHland FL, from step 2b. Lines U and L are which the hypothetical distribution will lie obtained by adding d to and subtracting d (at the stated i-confidence level). This is a from line S, respectively; d w 29% from Table very good approach, especially when the 2-12 for N w 20 and s-confidence -' 95%.
sample is plotted on special graph paper for The true distribution (uniform over zero to which the hypothetical distribution will be one) Is the 45-dog line from the origin to the a straight line. It Is explained more fully In point (1, 100%); it lies well within the s-con- par. 2-5. fidencc band bounded by the lines U and L, 100' 80I 40' 60 0 0.2 0.4 0.6 0.8 1.0 Random Variable W Fligui 2412 (Al. Kolmovorov-Smimov Limits (96% s.Confldence and Sample Cdt- from Table 2- 13)(