2.16 A researcher with a sample of 50 individuals with similar education but differing amounts of training hypothesizes that hourly earnings, EARNINGS, may be related to hours of training, TRAINING, according to the relationship EARNINGS =  1 +  2  TRAINING + u He is prepared to test the null hypothesis H 0 :  2 = 0 against the alternative hypothesis H 1 :  2 0 at the 5 percent and 1 percent levels. What should he report 1.If b 2 = 0.30, s.e.(b 2 ) = 0.12? 2.If b 2 = 0.55, s.e.(b 2 ) = 0.12? 3.If b 2 = 0.10, s.e.(b 2 ) = 0.12? 4.If b 2 = -0.27, s.e.(b 2 ) = 0.12? EXERCISE

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom There are 50 observations and 2 parameters have been estimated, so there are 48 degrees of freedom. 2 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 The table giving the critical values of t does not give the values for 48 degrees of freedom. We will use the values for 50 as a guide. For the 5% level the value is 2.01, and for the 1% level it is The critical values for 48 will be slightly higher. 3 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 1.If b 2 = 0.30, s.e.(b 2 ) = 0.12? t = In the first case, the t statistic is EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 1.If b 2 = 0.30, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 5% level but not at the 1% level. 5 This is greater than the critical value of t at the 5% level, but less than the critical value at the 1% level. EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 1.If b 2 = 0.30, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 5%, but not at the 1%, level. In this case we should mention both tests. It is not enough to say "Reject at the 5% level", because it leaves open the possibility that we might be able to reject at the 1% level. 6 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 1.If b 2 = 0.30, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 5%, but not at the 1%, level. Likewise it is not enough to say "Do not reject at the 1% level", because this does not reveal whether the result is significant at the 5% level or not. 7 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 2.If b 2 = 0.55, s.e.(b 2 ) = 0.12? t = In the second case, t is equal to EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 2.If b 2 = 0.55, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 1% level. We report only the result of the 1% test. There is no need to mention the 5% test. If you do, you reveal that you do not understand that rejection at the 1% level automatically means rejection at the 5% level, and you look ignorant. 9 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 2.If b 2 = 0.55, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 0.1% level (t crit, 0.1% = 3.50). Actually, given the large t statistic, it is a good idea to investigate whether we can reject H 0 at the 0.1% level. It turns out that we can. The critical value for 50 degrees of freedom is So we just report the outcome of this test. There is no need to mention the 1% test. 10 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 2.If b 2 = 0.55, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 0.1% level (t crit, 0.1% = 3.50). Why is it a good idea to press on to a 0.1% test, if the t statistic is large? Try to answer this question before looking at the next slide. 11 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 2.If b 2 = 0.55, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 0.1% level (t crit, 0.1% = 3.50). The reason is that rejection at the 1% level still leaves open the possibility of a 1% risk of having made a Type I error (rejecting the null hypothesis when it is in fact true). So there is a 1% risk of the "significant" result having occurred as a matter of chance. 12 EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 2.If b 2 = 0.55, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 0.1% level (t crit, 0.1% = 3.50). 13 If you can reject at the 0.1% level, you reduce that risk to one tenth of 1%. This means that the result is almost certainly genuine. EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 3.If b 2 = 0.10, s.e.(b 2 ) = 0.12? t = In the third case, t is equal to EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 3.If b 2 = 0.10, s.e.(b 2 ) = 0.12? t = Do not reject H 0 at the 5% level. 15 We report only the result of the 5% test. There is no need to mention the 1% test. If you do, you reveal that you do not understand that not rejecting at the 5% level automatically means not rejecting at the 1% level, and you look ignorant. EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 4.If b 2 = -0.27, s.e.(b 2 ) = 0.12? t = In the fourth case, t is equal to EXERCISE 2.16

EARNINGS =  1 +  2  TRAINING + u H 0 :  2 = 0, H 1 :  2 0 n = 50, so 48 degrees of freedom t crit, 5% = 2.01, t crit, 1% = 2.68 _______________________________________________ 4.If b 2 = -0.27, s.e.(b 2 ) = 0.12? t = Reject H 0 at the 5% level but not at the 1% level. The absolute value of the t statistic is between the critical values for the 5% and 1% tests. So we mention both tests, as in the first case. 17 EXERCISE 2.16

Copyright Christopher Dougherty 1999–2006. This slideshow may be freely copied for personal use