Exam1_5507_sol

# Exam1_5507_sol - STA5507 Section 8112 EXAM 1. Solutions 1....

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Unformatted text preview: STA5507 Section 8112 EXAM 1. Solutions 1. Below are recorded the white blood cell counts for a sample of 6 adult males with leukemia (recorded in thousands of cells). 19.3 32.0 1.8 43.0 14.1 1.4 Making no assumptions about the distribution of white cell counts in this population, find a point estimate of the median white blood cell count for the population of adult male with leukemia. Also, fine roughly 80% confidence intervals for the population median cell count and interpret it. (10 pts) ˆ θ . 5 = 14 . 1 + 19 . 3 2 = 16 . 7 From R , a roughly 80% confidence interval for the population median is given by [ X (2) ,X (5) ) = [1 . 8 , 32 . 0) and the achieve confidence level is 4 summationdisplay k =2 parenleftbigg 6 k parenrightbiggparenleftbigg 1 2 parenrightbigg 6 = 0 . 78125 . We are 78 . 13% confident that the median white blood cell count for the population of adult male with leukemia falls in the interval [1 . 8 , 32 . 0). 2. Let X ( 1) ≤ X ( 2) ≤ ··· ≤ X ( n ) be the order statistics from a random sample of exponential distribution with λ = 1. In other words, X 1 ,... ,X n ∼ exp ( λ = 1) . Thus, the distribution function of X has the form of f ( x ) = e- x x > . Show that E bracketleftbig X ( j ) bracketrightbig = j- 1 summationdisplay i =0 1 n − i = j summationdisplay i =1 1 n − i + 1 . Hints are given below (10 pts) E [ X ] = integraldisplay ∞ [1 − F ( t )] d t B ( a,b ) = integraldisplay 1 u a- 1 (1 − u ) b- 1 d u = Γ( a )Γ( b ) Γ( a + b ) Beta function Γ( x ) = ( x − 1)! if x is an integer Gamma function For a exponential distribution with λ = 1, the cdf is given by F ( t ) = Pr ( X ≤ t ) = integraldisplay t f ( x )d x = integraldisplay t e- x d x = 1 − e- t ( t > 0) Thus, F X ( j ) ( t ) = Pr ( X ( j ) ≤ t ) = Pr [at lest j on the X ’s ≤ t ] = n summationdisplay l = j Pr (exactly l of the X ’s ≤ t ) = n summationdisplay l = j parenleftbigg n l parenrightbigg [ F ( t )] l [1 − F ( t )] n- l = n summationdisplay l = j parenleftbigg n l parenrightbigg ( 1 − e- t ) l braceleftbig 1 − ( 1 − e- t )bracerightbig n- l = n summationdisplay l = j parenleftbigg n l parenrightbigg ( 1 − e- t ) l ( e- t ) n- l E bracketleftbig X ( j ) bracketrightbig = integraldisplay ∞ bracketleftBig 1 − F X ( j ) ( t ) bracketrightBig d t = integraldisplay ∞ 1 − n summationdisplay l = j parenleftbigg n l parenrightbigg ( 1 − e- t ) l ( e- t ) n- l d t = integraldisplay ∞ j- 1 summationdisplay l =0 parenleftbigg n l parenrightbigg...
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## This note was uploaded on 07/22/2011 for the course STA 4502 taught by Professor Staff during the Spring '08 term at University of Florida.

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Exam1_5507_sol - STA5507 Section 8112 EXAM 1. Solutions 1....

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