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Statistical Methods in Economics


                   Notes                                              15
                                                                =  67  +  ×  3  = 64 + (0.45) = 67.45
                                                                      100
                                              (b) The unbiased and efficient estimate of the population variance is:
                                                                   n   2
                                                             ˆ s 2  =   n  − 1  ⋅s


                                                                  Σ  2  ⎛ fd  Σ '  ' ⎞ fd  2
                                                 where,      s 2  =   −  ⎜  ⎝ n  ⎟  ⎠ n  ×  i 2


                                                                  ⎡    ⎛   ⎞ 15  2  ⎤ 97
                                                                =   ⎢  −  ⎜  ⎟  ⎥  ×  3 2
                                                                  ⎢    ⎝ 100  ⎠ 100  ⎦  ⎥ ⎣
                                                                = [0.97 – .0225] × 9 = 8.5275
                                                                   n   2  100
                                                 Now,        ˆ s 2  =   n  − 1 s  =   99  × 8.5275  = 8.6136


                                                 Thus,        μ = 67.45,   σ 2   = 8.6136
                                  (2)  Point Estimation in Case of Repeated Sampling: When large number of random samples of
                                      same size are drawn from the population with or without replacement, then the point estimates
                                      of the population parameter can be illustrated by the following examples:
                                  Example 10:  A population consists of five values: 3, 4, 5, 6 and 7. List all possible samples of size 3
                                              without replacement from this population and calculate the mean   X   of each sample.
                                              Verify that sample mean   X   is an unbiased estimate of the population mean.

                                  Solution:   The population consists of the five values: 3, 4, 5, 6, 7. The total number of possible
                                                                                5
                                              samples of size 3 without replacement are  c  = 10 which are shown in the following
                                                                                 3
                                              table:
                                        Sample No.          Sample Values           Sample Mean ()
                                                                                                 X
                                            (1)                  (2)                       (3)
                                                                                    1           12
                                             1                 (3, 4, 5)             (  34  ) ++ 5  =    = 4
                                                                                    3           3

                                                                                    1           13
                                                                                            ) ++
                                             2                 (3, 4, 6)             (  346  =     = 4.33
                                                                                    3           3
                                                                                    1           14
                                             3                 (3, 4, 7)             (  34  ) ++ 7  =    = 4.67
                                                                                    3           3

                                                                                    1           14
                                             4                 (3, 5, 6)             (  35  ) ++ 6  =    = 4.67
                                                                                    3           3
                                                                                    1           15
                                             5                 (3, 5, 7)             (  35  ) ++ 7  =    = 5.0
                                                                                    3           3
                                                                                    1           16
                                                                                            ) ++
                                             6                 (3, 6, 7)             (  367  =     = 5.33
                                                                                    3           3



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