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Quantitative Techniques-II



                      Notes                                     2              2              2
                                                           x x    70   x x    62   x x    60
                                         Sample variance =                                   
                                                                     
                                                           n 1     5 1 ,  n 1     5 1 ,  n 1     6 1
                                                             
                                                                                                  
                                                                                           
                                                                            
                                                                                   
                                                             70             62             60
                                                          2
                                                                                        2
                                                                         2
                                                         s      17.5  s      15.5  s      12
                                                                         2
                                                          1
                                                                                        3
                                                             4              4              5
                                                               2     n   1   2
                                                                      i
                                    5.   Within column variance           s 1
                                                                        k
                                                                     n  
                                                                      i
                                                                                       
                                                             
                                                                          
                                                            5 1        5 1        6 1 
                                                        =         17.5        15.5        12
                                                           16 3        16 3       16 3 
                                                             
                                                                                        
                                                                           
                                                            4        4      5
                                                        =       17.5        15.5    12
                                                           13        13     13
                                                               192
                                         Within column variance      14.76
                                                                13
                                             Between column variance  20
                                    6.   F =                             1.354
                                             Within column variance  14.76
                                    7.   d.f of Numerator = (3 – 1) = 2.
                                    8.   d.f of Denominator =    n   k  = (5 – 1) + (5 – 1) + (6 – 1) = 16 – 3 = 13.
                                                              1
                                    9.   Refer to table using d.f = 2 and d.f = 13.
                                    10.  The value is 3.81. This is the upper limit of acceptance region. Since calculated value 1.354
                                         lies within it we can accept H0, the null hypothesis.
                                    Conclusion: There is no significant difference in the effect of the three training methods.
                                    Self Assessment
                                    Fill in the blanks:
                                    3.   For using ANOVA, the data should be ............................. in nature.

                                    4.   F test has ............................. parameters.
                                    13.3 Summary


                                        Testing the hypothesis about difference between two means: This can be used when two
                                         population means are given and null hypothesis is H  : P1 = P2.
                                                                                    o
                                        ANOVA is a statistical technique. It is used to test the equality of three or more sample
                                         means.  Based  on  the  means,  inference  is  drawn  whether  samples  belongs  to  same
                                         population or not.

                                    13.4 Keywords


                                    ANOVA: It is a statistical technique. It is used to test the equality of three or more sample means.
                                    Based on the means, inference is drawn whether samples belongs to same population or not.





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