ECO 329 - Dr. V.R. Bencivenga Economics 329 Fall 2010...

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Dr. V.R. Bencivenga September 29, 2010 Economics 329 Fall 2010 MIDTERM EXAM #1: RIGHT ANSWERS 1.a. 8 . 36 ) 35 45 )( 218 . 46 . 5 . ( 35 median years b. The median for the distribution based on all households is larger. It lies in the fourth class interval, because the cumulative relative frequency of the third class interval is 49.5%, and the cumulative relative frequency of the fourth class interval is 64.5%. Therefore, the median is between 45 and 55 . 2.a. K 1 k k k Y f Y In this problem, the Y k is the k th value in the data set, and f k is the number of repetitions of that value. Using this approach Y = .15(.1 + .2) + .20(.1 + .1 + .1 + .1) + .25(.2 + .1) = .15(.3) + .20(.4) + .25(.3) = .20 An alternative expression would be K 1 k k k Y n 100 1 Y Y = (1/100)[ 15(.1 + .2) + 20(.1 + .1 + .1 + .1) + 25(.2 + .1) ] = .20 b. K 1 k 2 k k 2 Y ) Y Y ( n 1 n 1 s 001515 . ] ) 0025 (. 30 ) 0025 (. 30 [ 99 1 ] ) 20 . 25 (. 30 ) 20 . 20 (. 40 ) 20 . 15 (. 30 [ 99 1 s 2 2 2 2 Y Note it is also acceptable to use the approximate formula : K 1 k 2 k k 2 Y ) Y Y ( f s 2 Y s = (.15 - .20) 2 (.3) + (.20 - .20) 2 (.4) + (.25 - .20) 2 (.3) = .0015 c. We first need X = 2(.3) + 3(.2) + 4(.2) + 5(.3) = 3.5 Cov(X,Y) = s XY = (1/99) [ (2 3.5)(.15 .20)(0) + (2 3.5)(.20 .20)(10) + (2 3.5)(.25 .20)(20) + (3 3.5)(.15 .20)(0) + (3 3.5)(.20 .20)(10) + (3 3.5)(.25 .20)(10) + (4 3.5)(.15 .20)(10) + (4 35)(.20 .20)(10) + (4 3.5)(.25 .20)(0) + (5 3.5)(.15 .20)(20) + (5 3.5)(.20 .20)(10) + (5 3.5)(.25 .20)(0) ]
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2 Note that there are only four non-zero terms . Some terms equal zero because there are no data in those “cells.” Other terms equal zero because either or both “deviations from the mean” equal zero. Units of the covariance are
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This note was uploaded on 06/15/2011 for the course ECO 329 taught by Professor K during the Fall '08 term at University of Texas at Austin.

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ECO 329 - Dr. V.R. Bencivenga Economics 329 Fall 2010...

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