Page 370 - Fundamentals of Probability and Statistics for Engineers
P. 370

Linear Models and Linear Regression                             353

                                                        1=2
                                    8                 9
                                       "           #  1
                                          n
                                    <                 =
                                         X        2
                              t n 2;
=2   b 2  …x i   x†  :             …11:44†
                                    :                 ;
                                         iˆ1
           Similarly, significance tests about the value of    can be easily carried out with
                 ^
           use of A  as the test statistic.
             An  important  special  case  of  the  above  is  the  test  of  H 0 :   ˆ  0 against
           H 1 :   6ˆ  0. This particular situation corresponds essentially to the significance
           test of linear regression. Accepting H 0 is equivalent to concluding that there is
           no  reason  to  accept  a  linear  relationship  between  E  Y   and  x  at  a  specified
                                                        f g
           significance level 
 . In many cases, this may indicate the lack of a causal
                              f g
           relationship between E  Y   and independent variable x.
             Example 11.5. Problem: it is speculated that the starting salary of a clerk is a
           function of the clerk’s height. Assume that salary (Y ) is normally distributed and
           its mean is linearly related to height (x); use the data given in Table 11.3 to test
                             f g
           the assumption that E  Y   and x are linearly related at the 5% significance level.

                     Table 11.3 Salary, y (in $10 000), with height, x (in feet),


                                     for Example 11.5
                     x   5.7  5.7   5.7  5.7   6.1  6.1   6.1   6.1
                     y   2.25  2.10  1.90  1.95  2.40  1.95  2.10  2.25
             Answer: in this case, we wish to test H 0 :   ˆ  0 against H 1 :   6ˆ  0, with 
 ˆ  0:05.
             From the data in Table 11.3, we have

                                                               1
                                 n                 X
                                                    n
                               "                #"           #
                            ^
                                X
                             ˆ     …x i   x†…y i   y†  …x i   x† 2
                                 iˆ1               iˆ1
                             ˆ 0:31;
                       t n 2;
=2 ˆ t 6;0:025 ˆ 2:447;  from Table A.4;
                                    "                         #
                                      n              n
                                 1   X        2     X        2
                           b 2
                             ˆ          …y i   y†     ^ 2  …x i   x†  ˆ 0:02;
                               n   2
                                     iˆ1            iˆ1
                   n
                  X         2
                     …x i   x† ˆ 0:32:
                  iˆ1
             According to Equation (11.44), we have
                                                    1=2
                                8                  9
                                    "           #  1
                                      n
                                <                  =
                                     X         2
                           t 6;0:025   b 2  …x i   x†  ˆ 0:61:
                                :                  ;
                                      iˆ1


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