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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online