SW_2e_ex_ch03[1] - Chapter 3 Review of Statistics Review...

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Chapter 3 Review of Statistics R eview the Concepts ± Solutions to Exercises 1. The central limit theorem suggests that when the sample size ( n ) is large, the distribution of the sample average ( Y ) is approximately 2 , Y Y N µσ    with 2 2 . Y n Y σ = Given a population 100, Y µ = 2 43 0, Y =. we have (a) 100, n = 2 2 43 100 043 , Y n Y === . and 100 101 100 Pr( 101) Pr (1.525) 0 9364 Y Y −− <= < Φ = .. (b) 64, n = 2 2 43 64 64 06719 , Y Y . and 101 100 100 103 100 Pr(101 103) 0 6719 0 6719 0 6719 (3 6599) (1 2200) 0 9999 0 8888 0 1111 Y Y << = < < ... ≈Φ . −Φ. −. . (c) 165, n = 2 2 43 165 0 2606, Y n Y . and 100 98 100 98) 1 Pr ( 98) 1 Pr 0 2606 0 2606 1 ( 3 9178) (3 9178) 1 0000 (rounded to four decimal places) Y YY >= ≤= ≈−Φ−. =Φ . . 2. Each random draw i Y from the Bernoulli distribution takes a value of either zero or one with probability Pr( 1) i Yp == and Pr( 0) 1 . i ==− The random variable i Y has mean ( ) 0 Pr( 0) 1 Pr( , i EY Y Y p = +× = = and variance 2 22 var( ) [( ) ] ( 0 )P r ( 0 ) ( 1 )P r ( 1 ) (1 ) ) ) ii Y YE Y pY pp p p p p =− =− × =+− × = + = .
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Solutions to Exercises in Chapter 3 15 (a) The fraction of successes is 1 (1 ) (success) ˆ n ii i Y #Y # pY nn n = = == = = . (b) 1 11 ˆ () n i Y Ep E EY p p n =  = = .   ∑∑ (c) 1 22 ( 1 ) ˆ var( ) var var( ) ) n i Y pp p p = === = . The second equality uses the fact that 1 Y , , Y n are i.i.d. draws and cov( , ) 0, ij YY = for . 3. Denote each voter’s preference by . Y 1 Y = if the voter prefers the incumbent and 0 Y = if the voter prefers the challenger. Y is a Bernoulli random variable with probability Pr( 1) Yp and Pr( 0) 1 . ==− From the solution to Exercise 3.2, Y has mean p and variance ). (a) 215 400 ˆ 0 5375. p . (b) · ˆˆ (1 ) 0.5375 (1 0.5375) 4 400 ˆ var( ) 6 2148 10 . n p −× = . × The standard error is SE 1 2 ( ) (var( )) 0 0249. . (c) The computed t -statistic is 0 ˆ 05375 05 1506 ˆ SE( ) 0 0249 p act p t p µ , .− . = . . . Because of the large sample size ( 400), n = we can use Equation (3.14) in the text to get the p -value for the test 0 05 Hp := . vs. 1 05: :≠ . -value 2 ( | |) 2 ( 1 506) 2 0 066 0 132 act pt =Φ− =Φ−. = ×. =. . (d) Using Equation (3.17) in the text, the p -value for the test 0 . vs. 1 :>. is -value 1 ( ) 1 (1 506) 1 0 934 0 066 act =−Φ =−Φ. =−. . (e) Part (c) is a two-sided test and the p -value is the area in the tails of the standard normal distribution outside ± (calculated t -statistic). Part (d) is a one-sided test and the p -value is the area under the standard normal distribution to the right of the calculated t -statistic. (f) For the test 0 . vs. 1 , :> . we cannot reject the null hypothesis at the 5% significance level. The p -value 0.066 is larger than 0.05. Equivalently the calculated t -statistic . is less than the critical value 1.645 for a one-sided test with a 5% significance level. The test suggests that the survey did not contain statistically significant evidence that the incumbent was ahead of the challenger at the time of the survey.
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16 Stock/Watson - Introduction to Econometrics - Second Edition 4. Using Key Concept 3.7 in the text (a) 95% confidence interval for p is ˆˆ 1.96 ( ) 0.5375 1.96 0.0249 (0.4887,0.5863).
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SW_2e_ex_ch03[1] - Chapter 3 Review of Statistics Review...

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