Mathematical Methods for Physics and Engineering - Matematica.NET

STATISTICSwhere α is the required significance level of the test. In our case we set α =0.05, and fromtable 31.3 with n = 16 we find that t crit =2.12. The rejection region is thereforet2.12.Since t = −2.17 for our samples, we can reject the null hypothesis H 0 : µ 1 = µ 2 , althoughonly by a small margin. (Indeed, it is easily shown that one cannot reject H 0 at the 2%significance level). The 95% central confidence interval on ω = µ 1 − µ 2 is given by( ) 1/2 ( ) 1/2 N1 + N 2N1 + N 2¯w − ˆσt crit < ω < ¯w + ˆσt crit ,N 1 N 2 N 1 N 2where t crit is given by (31.120). Thus, we find−26.1