Christopher Dougherty EC220 - Introduction to econometrics (chapter 2) Slideshow: exercise 2.16 Original citation: Dougherty, C. (2012) EC220 - Introduction to econometrics (chapter 2). [Teaching Resource] © 2012 The Author This version available at: Available in LSE Learning Resources Online: May 2012 This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 License. This license allows the user to remix, tweak, and build upon the work even for commercial purposes, as long as the user credits the author and licenses their new creations under the identical terms

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

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