Page 249 - DMGT209_QUANTITATIVE_TECHNIQUES_II
P. 249

Nourishment programme        Nourishment programme B
                                                                  A
                                           X            x x              2      y      y y               2
                                                         
                                                                                        
            Quantitative Techniques-II                               x x                          y y  
                                                       = (x-46)                       =(y-57)
                                           44            -2            4        42      -15           225
                                           37            -9            81       42      -15           225
                      Notes
                                           48            2             4        58      1              1
                                           60            14           196       64      7             49
                                           41            -5            25       64      7             49
                                                                                67      10            100
                                                                                62      5             25
                                           230           0            310      399      0             674

                                                                          
                                                                         x y
                                                                t =
                                                                          1  1 
                                                                      s 2      
                                                                          n  1  n 2 
                                    Here                       n  = 5                     n  = 7
                                                                 1                         1
                                                               x = 230                      y   399


                                                           x x   2  = 310                y    y  2    674

                                                                      x  230
                                                                x =   n    5    46
                                                                       1

                                                                      y   399
                                                                y =            57
                                                                      n    7
                                                                       2
                                                                         1            2         
                                                                                               2
                                                               s 2  =  n   n   2     x    x     y   y
                                                                      1
                                                                          2
                                                              D.F = (n +n -2) = (5 + 7-2) = 10
                                                                      1  2
                                                                      1
                                                               s 2  =  310 674     98.4
                                                                     10
                                                                         46 57
                                                                           
                                                                               1
                                                                t =   98.4     1   
                                                                             5  7 

                                                                          11
                                                                  =   98.4     12  
                                                                             35 


                                                                        11   11
                                                                  =         
                                                                      33.73   5.8
                                                                  = 1.89
                                    t at 10 d.f. at 5% level is 1.81.
                                    Since, calculated t is greater than 1.81, it is significant. Hence HA is accepted. Therefore the two
                                    nutrition programmes differ significantly with respect to weight increase.





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