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CPS1917 Trijya S.
                  3.   Result
                      Result In order to compare the performance of the proposed method to
                  the one proposed by Shah (1973) we have used the data reported in his paper.
                  The model being considered is as follows. After administering a dose of the
                                                                                        (10)
                  drug, its concentrations ( ), ( )..., ( ) are measured by taking a sample
                                                          
                                                   1
                                             0
                  from the central compartment at times  ,  , . . . ,  . For this sample, the model
                                                           1
                                                                 
                                                         0
                  could be written as
                                        ( ) =  −     +  −     +   ,
                                           
                                                                      
                  where   =  0, 1, . . . , .  We  assume  that    ’s  are  independently  distributed
                                                           
                  with ( ) = 0 and (  ) =   for all . The data set in Shah (1973) can
                                                   2
                                        
                  be described as follows. A panel of 8 subjects was administered 250 milligrams
                  (mg) of the drug Tetracycline Hyrdochloride after a specified breakfast. The
                  concentration of the drug in blood stream was measured at different points
                  of time. The time and the lowest concentration for all time points are given in
                  the first two columns of Table 1. Using the last six points of the elimination
                  phase, we obtain the decay function as
                                     () =  0.2793  −0.1331(−15) ,   ≥  15.
                  Therefore,    =  0.2793,  =  0.1331  and   =   =  14.1349.  Using  the
                                           ̂
                                                                  ̂
                                                            ̂
                                            
                              
                  methodology given in the previous section, the estimates of the first four raw
                  moments are given by   =  8.1648,   =  123.4072,   =  2782.1156 and
                                                         2
                                          1
                                                                          3
                    =  83634.2720. The quadratic equation (2) for this data set turns out to be
                    4
                  59.5243  −  481.3506 +  257.2676  =  0 .  The  roots  of  this  equation  are
                           2 ̂
                                         ̂
                    =  7.5112 and    = 0.5754 since the smaller root is taken to be .
                                    ̂
                                                                                   ̂
                  ̂
                                     1
                   2
                                                                                   1
                      The first approach (Method 1) used to estimate A, B,  and  in (10) utilizes
                  only  the  first  three  moments  and  can  be  described  as  follows.  Given  the
                                                                                       1
                            ̂
                                    ̂
                  estimates    and  , the estimates of  and  are obtained as,  ̂  =    ̂    and
                                     2
                             1
                                                                                       1
                  ̂
                     =   1  . We know that  is the coefficient of the first term of (9). Therefore, 
                         ̂
                         2
                                            −1
                                          
                                                        
                                                   
                                    
                  is the estimate of  (  +  ) . But   +   =   so given the estimate  we can
                                                                                    ̂
                                                
                                       
                  find the estimate of   as   and the estimate of A as  = ̂ . Using the
                                            ̂
                                                                                 ̂
                                                                         ̂
                                       
                              
                                                                
                  estimate of   , we can obtain the estimate of   as (1 −  ), and hence the
                                                                            ̂
                                                              
                  estimate of B is given by   = (1  −  ).
                                                 ̂
                                           ̂
                                                          ̂
                      The second approach (Method 2) uses a two stage procedure of estimation
                  which also utilizes the first four moments. In the first stage we estimate  
                                                                                           ̂
                                                                                            1
                  and     using  the  cubic  equation  (8),  which,  in  this  case  is  given  by
                       ̂
                        2
                    −  60.7012  +  31.3279  =  0 .  Since  4 + 27  = −867125.8  <  0 ,  all
                   3
                                                                    2
                                                            3
                   1
                  roots are real. Two  roots are given by estimate    and   are obtained as
                                                                    ̂
                                                                           ̂
                                                                            2
                                                                    1
                  0.4835 and 7.5190 which are then used to obtain estimates ̂ and . In the
                                                                                     ̂
                  second stage, given the estimates ̂ and  we transform the data as follows.
                                                           ̂
                  We       define       () =  (−̂) and  ()  =  (− ).   Using
                                                                              ̂
                                        1
                                                               2
                  {( ),   ( ),   ( )},   =  0, 1, . . . , ,  we  regress  ( )  on    ( )  and
                                                                                     
                          1
                             
                                                                         
                                    
                      
                                 2
                                                                                   1
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