"WEIBULL, CONVENTIONAL REGRESSION" [1] "beta=" "0.553" "0.707" "1.888" "2.863" [1] "sigma=" "0.9" [1] "lambda= 1" [1] "-2 Log Likelihood=" "4255.54" The difference with table 3 is 290.04= 2*sum(log(uncensored exits)) so that 4255.54+290.04= 4545.6