Metu department of economics instructor dr ozan

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METU - Department of Economics Instructor: Dr. Ozan ERUYGUR e-mail: [email protected] Lecture Notes 6 0 1 1 : 0.570 : 0.570 A H H 1 1 1 ˆ 0.724 0.570 2.2 ˆ 0.07 ( ) t se /2, 1 0.025,13 1 1 0.025,11 2.201 T k t t t     Since 0.025,11 2.2 2.201 t t , we do not reject null hypothesis that 1 0.570 at 0.05 level of significance. b) Now test 1 0.569 at 0.05 level. 0 1 1 : 0.569 : 0.569 A H H 1 1 1 ˆ 0.724 0.569 2.214 ˆ 0.07 ( ) t se Since 0.025,11 2.214 2.201 t t , we reject null hypothesis that 1 0.569 at 0.05 level of significance.
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ECON 301 - Introduction to Econometrics I April, 2012 METU - Department of Economics Instructor: Dr. Ozan ERUYGUR e-mail: [email protected] Lecture Notes 7 c) Test 1 0.8780 at 0.05 level. 0 1 1 : 0.8780 : 0.8780 A H H 1 1 1 ˆ 0.724 0.8780 2.2 ˆ 0.07 ( ) t se   Since 0.025,11 2.2 2.201 t t , we do not reject null hypothesis that 1 0.8780 at 0.05 level of significance. d) Test 1 0.8781 at 0.05 level. 0 1 1 : 0.8781 : 0.8781 A H H 1 1 1 ˆ 0.724 0.8781 2.2014 ˆ 0.07 ( ) t se   Since 0.025,11 2.2014 2.201 t t , we reject null hypothesis that 1 0.8781 at 0.05 level of significance. e) Obtain the confidence interval of 1 .
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ECON 301 - Introduction to Econometrics I April, 2012 METU - Department of Economics Instructor: Dr. Ozan ERUYGUR e-mail: [email protected] Lecture Notes 8 /2, 1 /2, 1 ˆ ˆ ˆ ˆ . ( ) . ( ) 1 i T k i i i T k i P t se t se     /2, 1 0.025,13 1 1 0.025,11 2.201 T k t t t     1 0.724 2.201(0.07) 0.724 2.201(0.07) 0.95 P 1 0.724 0.154 0.724 0.154 0.95 P 1 0.570 0.8780 0.95 P Example 2 Suppose we have estimated the following regression line from a sample of 22 observations where 0 ˆ ( ) se =38.2 and 1 ˆ ( ) se =0.85 ˆ 128.5 2.88 t t Y X (75.5) (0.21) From the t table, we see that the value of 0.025 t with 20 degrees of freedom is 2.086. Hence the 95 percent confidence interval for the parameters is: (1) For 0 0 128.5 2.086.(38.2) 128.5 79.69 or 0 128.5 79.69 128.5 79.69 0.95 P (1) For 1 1 2.88 2.086.(0.85) 2.88 1.77 or
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