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Table 1.1: Confidence Intervals for median using three bootstrap methods

lower bound upper bound

Normal Bootstrap CI 35.19 52.27Percentile Bootstrap CI 35.95 52.97Basic Bootstrap CI 34.78 51.81

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1.3 R Code

1 N=10002 n=503 cpu=s can (f i l e ="cp u tim e . txt " ) # IMPORT CPU DATA

4 ml. bs .median=r ep(0 ,N) # INIT ARRAY5 c i . normal=a r r a y( 0 , dim=c ( n , 2 ) ) # i n i t i a l i z e6 c i . p e r c e n t i l e =a r r a y( 0 , dim=c ( n , 2 ) ) # i n i t i a l i z e7 c i . b a s i c =a r r a y( 0 , dim=c ( n , 2 ) ) # i n i t i a l i z e8

9 n e g . l o g l i k . f u n = f u n c t i o n( pa r, da t ) { # FUNCTION USED IN OPTIM COMMAND FORMLE

10 i f ( par [ 2 ] = = 0 . 0 ) {11 par [ 2 ] = 0 . 0 0 0 0 0 0 0 0 1 # HACK TO PREVENT OPTIM ERRORS12 }13 r e s u l t = sum (dgamma ( dat , shape=par[ 1 ] , r a t e =par[ 2 ] ,l o g=TRUE) )14 r e t u r n ( r e s u l t )15 }16

17 m l . es t=optim ( par=c(3 ,0 .1 ) , fn=neg . l o g l i k . fun , method="LBFGSB",l o w e r=r ep( 0 , 2) , he ss i an=TRUE, dat=cpu ) # MLE GAMMA PARAMETER ESTIMATES

18 ml .median=qgamma ( 0 . 5 , m l . e s t $p ar[ 1 ] , ml . e s t $p ar[ 2 ] ) # MLE MEDIAN ESTIMATE OFCPU POPULATION

19 f o r ( i i n 1 :n ) {20 f o r ( i i i n 1 :N ) {21 bs=sample ( cpu ,l e n g t h(cp u ) ,r e p l a c e=T) # RESAMPLE CPU DATA22 ml. bs . es t=optim ( par=c(3 , 0. 1) , fn=neg . l o g l i k . fun , method="LBFGSB",

l o w e r=r ep( 0 ,2 ) , h es si an=TRUE, dat=bs) # MLE OF GAMMA PARAMETERS FROMBOOTSTRAP SAMPLE

23 ml. bs .median[ i i ]=qgamma (0 .5 , ml . b s . es t $p ar[ 1 ] , ml . b s . e s t $p ar[ 2 ] ) #BOOSTRAP ESTIMATE OF MEDIAN

24 }25

26 boot . bi as=mean (ml . bs .median )ml .median # BIAS27 boot .s e=sd (ml . bs .median ) # STANDARD ERROR28

29 # USING DEVELOPED FORMULAS I N CLASS CONFIDENCE INTERVALS USING 3BOOTSTRAP METHOD FOR CONFIDENCE INTERVALS

30 c i . normal [ i ,] = c ( m l .medianboot . biasqnorm ( 0 . 9 7 5 ) boot .s e, m l .medianboot. b iasqnorm ( 0 . 0 2 5 ) boot .s e) #NORMAL

31 c i . p e r c e n t i l e [ i , ] =s o r t(ml . bs .median ) [ c (c e i l i n g(N . 0 5 / 2) ,c e i l i n g(N(10.05 / 2 ) ) ) ] # PERCENTILE

32 c i . b as ic [ i , ]= c ( 2ml .medianc i . p e r c e n t i l e [ i , 2 ] , 2 ml .medianc i . p e r c e n t i l e[ i , 1 ] ) # BASIC

33 }34

35 # AVERAGE IN n=50 REALIZATIONS36 p r i n t( c ( mean ( ci . n or mal [ , 1 ] ) , mean ( ci . nor m al [ ,2 ] ) ))37 p r i n t( c ( mean ( c i . p e r c e n t i l e [ , 1 ] ) , mean ( c i . p e r c e n t i l e [ , 2 ] ) ) )38 p r i n t( c ( mean ( c i . b a s i c [ , 1 ] ) , mean ( c i . b a s i c [ , 2 ] ) ) )

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