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handout4_1 - Econ 139: Introduction to Econometrics Andrew...

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Econ 139: Introduction to Econometrics Andrew Sweeting 1 Department of Economics Duke University Spring 2011 Econ 139 Handout 4 (Duke) Hypothesis Testing Spring 2011 1 / 47
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mind. e.g., did most Duke faculty vote for Obama? sample voted for Obama what can we conclude? is this consistent with a hypothesis that most faculty supported Obama? to do this we will use our knowledge of the estimator±s distribution to construct and perform hypothesis tests . Econ 139 Handout 4 (Duke) Hypothesis Testing Spring 2011 2 / 47
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we expect the population parameter to lie with some probability. E.g., Pr ( ¯ X error < μ X < ¯ X + error ) = 0 . 95 0.95 is the standard probability used and re±ects the fact that we want to be conservative (i.e., wide intervals around our estimate). Econ 139 Handout 4 (Duke) Hypothesis Testing Spring 2011 3 / 47
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Let±s begin with a closely related exercise. Find an interval that will contain a random draw from a standard normal (which has mean zero) distribution 95% of the time From our previous analysis and the standard normal tables we know Pr ( 1 . 96 < Z < 1 . 96 ) = 0 . 95 3 1 0 -1 -3 0.95 0.025 0.025 1.96 -1. 96 There is a 95% chance that the random variable Z will fall in the interval ( 1 . 96 , 1 . 96 ) Econ 139 Handout 4 (Duke) Hypothesis Testing Spring 2011 4 / 47
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μ X based on the sample mean estimate ¯ X From the CLT we know that ¯ X μ X σ X p n is (approximately) N ( 0 , 1 ) . Therefore Pr 1 . 96 < ¯ X μ X σ X p n < 1 . 96 ± = 0 . 95 as above. Econ 139 Handout 4 (Duke) Hypothesis Testing Spring 2011 5 / 47
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Pr 1 . 96 < ¯ X μ X σ X p n < 1 . 96 ! = 0 . 95 Rearranging, Pr ¯ X 1 . 96 ± σ X p n < μ X < ¯ X + 1 . 96 ± σ X p n ± = 0 . 95 . μ X is ¯ X ² 1 . 96 ± σ X p n = ² ¯ X 1 . 96 ± σ X p n , ¯ X + 1 . 96 ± σ X p n ³ Notice that the interval gets tighter as n Econ 139 Handout 4 (Duke) Hypothesis Testing Spring 2011 6 / 47
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Example magic tells us that σ 2 = 16 Pr ¯ X 1 . 96 ± σ X p n < μ X < ¯ X + 1 . 96 ± σ X p n ± = 0 . 95 . Pr 20 . 5 1 . 96 ± 4 10 < μ X < 20 . 5 + 1 . 96 ± 4 10 ± = 0 . 95 ) ² 20 . 5 1 . 96 ± 4 10 , 20 . 5 + 1 . 96 ± 4 10 ³ = [ 19 . 7 , 21 . 3 ] Econ 139 Handout 4 (Duke) Hypothesis Testing Spring 2011 7 / 47
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Interpretation ¯ X 1 . 96 ± σ X p n = ¯ X ² 1 . 96 ± σ X p n , ¯ X + 1 . 96 ± σ X p n ± Which statement about the interval above is correct? 1
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This note was uploaded on 08/02/2011 for the course ECON 139 taught by Professor Alessandrotarozzi during the Spring '08 term at Duke.

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handout4_1 - Econ 139: Introduction to Econometrics Andrew...

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