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Statistics



                      Notes
                                           Example 4: The standard deviation of a random sample of size 81 was found to be 12. Test
                                    the hypothesis that population standard deviation is greater than 10.
                                    Solution.
                                    We have to test H  : s £ 10 against H  : s > 10.
                                                   0             a
                                                         (12 10-  )
                                                                    ´
                                                      z =          2 81 =  2.55.
                                                            10
                                    Since this value is greater than 1.645, H  is rejected. Hence, the sample information supports the
                                                                    0
                                    contention that s is greater than 10.
                                    32.3 Test of Hypothesis Concerning the Equality of Standard
                                          Deviations (Small Samples)


                                                                                                    s 2  /s 2
                                    We have to test H  : s  = s  against s  > s . Refer to § 20.6, the statistic  F =  1  1  , would
                                                   0  1   2        1   2                             2   2
                                                                                                    s 2  /s 2
                                            s 1 2
                                    become   2   under H , follows F - distribution with  n  (= n  - 1) and  n  (= n  - 1) degrees of
                                            s         0                          1   1         2   2
                                            2
                                    freedom.
                                    Remarks:

                                                                                                          2
                                                           1              2    n 1  2    1   æ    2 å   X ö
                                                                                                          1i
                                                     2
                                                    s =
                                    1.   We can write  1      å  (X -  X 1  ) =    S =       çå X -         ÷  and
                                                                                                  1i
                                                                                    1
                                                                   1i
                                                         n - 1               n -  1    n -  1           n
                                                          1                   1         1    è           1  ø
                                                                                                2
                                                1              2    n    2     1  æ     2 å   X ö
                                                                                                2i
                                          2
                                         s =       å  (X -  X  ) =   2  S =       çå  X -
                                          2             2i    2          2              2i        ÷ .
                                              n - 1                n -  1    n -  1  è       n 2  ø
                                                                              2
                                                                    2
                                               2
                                                               s 2
                                    2.   In the variance ratio  F =  1 2  ,  we take, by convention the largest of the two  sample
                                                               s
                                                                2
                                                    2
                                         variance as s . Thus, this test is always a one tailed test with critical region at the right
                                                    1
                                         hand tail of the F - curve.
                                                                                      s 2
                                    3.   The 100(1 - a)% confidence limits for the variance ratio   1 2  ,  are given by
                                                                                      s 2
                                            s é  2  1  s 2  s 2  1  ù
                                                                         -
                                         P ê  1  ×    1    1  ×   ú  =  1 a .
                                             2          2   2
                                            s ê ë  2  F a /2  s 2  s 2  F 1 a /2 ú û
                                                                -
                                           Example 5: Two independent samples of sizes 10 and 12 from two normal populations
                                    have their mean square deviations about their respective means equal to 12.8 and 15.2 respectively.
                                    Test the equality of variances of the two populations.
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