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Unformatted text preview: Problem Set 3 ECON 231W Spring 2010 Suggested Solutions 1. From the definition of covariance: XY = 1 n n X i =1 ( X i X )( Y i Y ) This is identical to: XY = 1 n n X i =1 ( X i X ) Y i ( X i X ) Y (1) and XY = 1 n n X i =1 ( Y i Y ) X i ( Y i Y ) X (2) Equation (1) can be rewritten as: XY = 1 n n X i =1 ( X i X ) Y i 1 n n X i =1 ( X i X ) Y = 1 n n X i =1 ( X i X ) Y i Y 1 n n X i =1 ( X i X ) = 1 n n X i =1 ( X i X ) Y i Y = 1 n n X i =1 ( X i X ) Y i Analogously for equation (2): XY = 1 n n X i =1 ( Y i Y ) X i 1 n n X i =1 ( Y i Y ) X = 1 n n X i =1 ( Y i Y ) X i X 1 n n X i =1 ( Y i Y ) = 1 n n X i =1 ( Y i Y ) X i X = 1 n n X i =1 ( Y i Y ) X i 2. a. For the people with at least one parent who went to college, a confidence interval for yrsed c is given by 1 yrsed c 1 . 96 SE ( yrsed c ) We know that yrsed c = 14 . 8 and SE ( yrsed c ) = s c n c = 1 . 74 954 = . 056 . Therefore the confidence interval is yrsed c 1 . 96 SE ( yrsed c ) = 14 . 8 1 . 96 . 056 = (14 . 69 , 14 . 91) For the people for which neither parent went to college, a confidence interval for yrsed nc is given by yrsed nc 1 . 96 SE ( yrsed nc ) In this case, yrsed nc = 13 . 5 and SE ( yrsed nc ) = s nc n nc = 1 . 72 2842 = . 032 . Therefore the confidence interval is yrsed nc 1 . 96 SE ( yrsed nc ) = 13 . 5 1 . 96 . 032 = (13 . 44 , 13 . 56) b. The null and alternative hypotheses are written: H : c nc = 0 H A : c nc 6 = 0 To test this hypothesis we can either construct a confidence interval (using z / 2 = 2 . 58) or simply calculate the pvalue...
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This note was uploaded on 04/19/2010 for the course ECO 231W taught by Professor Joshuakinsler during the Spring '10 term at Rochester.
 Spring '10
 JoshuaKinsler

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