"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