Law School Data GPA LSAT score

Transcription

1 Bootstrap Estimation Suppose a simple ranom sample (sampling with replacement) X ; X ; : : : ; X n is available from some population with istribution function (cf) F (x): Objective: Make inferences about some feature of the population ffl meian ffl variance ffl correlation 08 A statistic t n is compute from the observe ata: Sample mean: t n = n Stanar eviation: t n = vu u t Correlation: t n = X j = 6 4 n X Pn j= X j n (X j X) j= X j X j j = ; ; : : : ; n Pn j= (X j X : )(X j X : ) vu u t P n j= (X j X : ) v u ut Pn j= (X j X : ) 08 t n estimates some feature of the population. What can you say about the istribution of t n, with respect to all possible samples of size n from the population? ffl Expectation of t n ffl Stanar eviation ffl Distribution function How can a confience interval be constructe? Simulation: (The population c..f. is known) ffl For a univariate normal istribution with mean μ an variance ff, the cf is F (x) = P rfx» xg = Z x p e (w μ) ff w ßff ffl Simulate B samples of size n an compute the value of t n for each sample: t n; ; t n; ; : : : ; t n;b

2 ffl Approximate E F (t n ) with the simulate mean t = B B X k= t n;k ffl Approximate V ar F (t n ) with B X B (t n;k t) k= ffl Approximate the stanar eviation of t n with vu u t B B X k= (t n;k t) ffl Approximate the c..f. with for t n F n (t) = number of samples with t n;k < t B ffl Orer the B values of t n smallest to largest t n()» t n()» : : :» t n(b) from an approximate percentiles of the istribution of t n What if F (X), the population c..f. is unknown? ffl You cannot use a ranom number generator to simulate samples of size n an values of t n, from the actual population. ffl Use a bootstrap (or resampling) metho? Basic iea: () Approximate the population cf F (x) with the empirical cf ^F n (x) obtaine from the observe sample X ; X ; : : : ; X n Assign probability to each observation in the sample. n Then F b n (x) = n where b is the number of observations in the sample with X i < x for i = ; ; : : : ; n

3 ffl Approximate the act of simulating B samples of size n from a population with c..f. F (x) by simulating B samples of size n from a population with c..f. F n (x) The approximating" population is the original sample. Sample n observations from the original sample using simple ranom sampling with replacement. This will be calle a boostrap sample. repeat this B times to obtain B bootstrap samples of size n. Evaluate the summary statistic for each bootstrap sample Sample : t Λ n; Sample : t Λ n;. Sample B: t Λ n;b ffl Evaluate bootstrap estimates of features of the sampling istribution for t n, when the c..f. is ^F n (x). V ar Fn (t n ) E Fn (t n ) = B = B ±B b= B X b= tλ n;b 4 t Λ n;b E F n (t n ) 3 5 ffl This resampling proceure is calle a nonparametric bootstrap ffl If use properly it provies consistent large sample results: As n! an B! E Fn (t n ) = B X B b= tλ n;b V ar Fn (t n ) =! E F (t n ) B B X 6 4t Λ n;b 3 B X B b= tλ 7 n;b5 b=! V ar F (t n ) 09 09

4 The bootstrap is a large sample metho ffl Large number of bootstrap samples. As B! ffl Consistency: Original sample size must become large As n! ; ^F n (x)! F (x) for any x E ^Fn (t n ) = B V ar ^Fn (t n ) = B X b= tλ n;b! E ^Fn (t n ) B B X! V ar ^Fn (t n ) b= (tλ n;b What is a goo value for B? stanar eviation: B ß 00 confience interval: B ß 000 more emaning applications: B ß 5000 E ^Fn (t n )) 093 Then, E ^Fn (t n )! E F (t n ) V ar ^Fn (t n )! V ar F (t n ).. ^F n (x) ffl For small values of n; coul eviate substantially from F (x). 094 Example.: Average values for GPA an LSAT scores for stuents amitte to n=5 Law Schools in 973. School LSAT GPA GPA Law School Data LSAT score

5 ffl These schools were ranomly selecte from a larger population of law schools. ffl We want to make inferences about the correlation between GPA an LSAT scores for the population of law schools ffl The sample correlation (n=5) is r = 0:7764 t n Bootstrap samples Take samples of n=5 schools, using simple ranom sampling with replacement Sample : School LSAT GPA t Λ n; = 0:8586 = r 5; Sample : School LSAT GPA Repeat this B = 5000 times to obtain t Λ n; ; tλ n; ; : : : ; tλ n;5000 t Λ n; = 0:6673 = r 5;

6 Estimate correlation from the original sample of n = 5 law schools 5000 Bootstrap Correlations r = 0: Bootstrap stanar error (from B = 5000 bootstrap samples) is S r = vu u t B B X 6 b= 4t Λ n;b B 3 B X tλ 7 n;j5 j= = vu u t X 6 b= 4r 5;b X r 7 5;j 5 j= = 0: Number of Bootstrap samples Bootstrap estimate of Stanar error B = 5 0:08 B = 50 0:0985 B = 00 0:334 B = 50 0:306 B = 500 0:38 B = 000 0:99 B = 500 0:366 B = :34 GPA Law School Data Very selom are more than B = 00 replications neee for estimating a stanar error." Efron & Tibshirani (993) (page 5) LSAT score 03 04

7 School LSAT GPA Type School LSAT GPA Type School LSAT GPA Type School LSAT GPA Type

8 The correlation coefficient for the population of 8 Law Schools in 973 is ρ = 0:7600 The exact istribution of estimate correlation coefficients for ranom samples of size n=5 from this population involves More than possible samples of size n=5 Results from 00,000 samples of size n=5. Percentiles min = max = mean = st. error = Bias: From a sample of size n, t n is use to estimate a population parameter. Bias F (t n ) = E F (t n ) " " average across true all possible parameter samples of size n value from the population Law School example: Bias F (r 5 ) = E F (r 5 ) 0:7600 = :7464 :7600 = 0:036 Bootstrap estimate of bias: true value" if you take the original sample as the population Bias F (t n ) = E F (t n ) t n " approximate this with the average of results from B bootstrap samples PB B b= t Λ n;b #

9 Improve bootstrap bias estimation: Efron & Tibshirani (993), Section 0.4 Law School example: (B = 5000 bootstrap samples) Bias F (r 5 ) = 5000 ±5000 b= r 5;b r 5 = :769 :7764 = :007 3 Bias correcte estimates: f t n = t n Bias b (t n ) = t n 0 B Law School example: ~r 5 = r 5 Bias b (r 5 ) B X tλ C n;ba b= = :7764 ( :007) = 0:7836 " here we move farther away from ρ = :7600: 4 Bias correction can be angerous in practice: ffl f t n may have a substantially larger variance than t n ffl MSE F ( f t n ) = [Bias F ( f t n )] + V ar F ( f t n ) is often larger than MSE F (t n ) = [Bias F (t n )] + V ar F (t n ) Empirical Percentile Bootstrap Confience Intervals Construct an approximate ( ff) 00% confience interval for. ffl t n is an estimator for ffl Compute B bootstrap samples to obtain t Λ n; ; : : : ; tλ n;b ffl Orer the bootstrappe values from smallest to largest t n;()» t n;()» : : :» t n;(b) 5 6

10 ffl Compute upper an lower 00th percentiles ff Compute k L = 6 4(B + ) ff = largest integer» (B + ) ff Law School example: (B = 5000 bootstrap samples) Construct an approximate 90% confience interval for ρ = population correlation ff = :0 k U = B + k L k L = [(500)(:05)] = [50:05] = 50 Then, an approximate ( ff) 00% confience interval for is [t n;(kl ); t n;(ku ) ] k U = = 475 An approximate large sample 90% confience interval is [r 5;(50) ; r 5;(475) ] = [0:507; 0:9487] Bootstrap Correlations Law School example: ffl Fisher Z-transformation Z n = 0 log + r n B _ο N log r n 0 C A + ρ ρ C A ; n 3 ffl An approximate ( ff) 00% confience interval for 0 +ρ A ρ is lower limit: Z n Z ff= p n 3 = Z L C A upper limit: Z n +Z ff= p n 3 = Z U ffl Transform back to the original scale 6 4 e Z L e ; ezu Z L + e Z u

11 For n = 5 an r 5 = :7764 we have an Z 5 = 0 log + :7764 C A = :0364 :7764 Z L = Z 5 Z :05 p 5 3 = :5637 Z U = Z 5 + Z :05 p 5 3 = :53 an an approximate 90% confience interval for the correlation is (0:509; 0:907) The bootstrap percentile interval woul approximate this interval if the original sample was taken from a bivariate normal approximation. Percentile Bootstrap Confience Intervals Suppose there is a transformation such that ffi = m( ) ffi = m(t n ) ο N(ffi;! ) for some stanar eviation!. Then, an approximate confience interval for is [m ( ffi Z (ff=)!); m ( ffi + Z ff=!)] ffl The bootstrap percentile interval is a consistent approximation. (You o not have to ientify the m() transformation. ffl The bootstrap approximation becomes more accurate for larger sample ffl For smaller samples, the coverage probability of the bootstrap percentile interval tens to be smaller than the nominal level ( ff) 00. ffl A percentile interval is entirely insie the parameter space. A percentile confience interval for a correlation lies insie the interval [ ; ]. 3 4

12 Bias-correcte an accelerate (BCa) bootstrap percentile confience intervals ffl simulate B bootstrap samples an orer the resulting estimates t Λ n;()» tλ n;()»» tλ n;(b) ffl The BC a interval of intene coverage ff is given by where ff ff [t Λ n;([ff(b+)]) ; tλ n;([ff(b+)]) ] = Φ = Φ 0 Z Z 0 + Z 0 Z ff= a( Z 0 Z ff= ) Z 0 + Z ff= a( Z 0 + Z ff= ) C A C A 5 6 an Φ() Z ff= Z 0 = Φ ψ is the c..f. for the stanar normal istribution is an upper" percentile of the stanar normal istribution, e.g., Z :05 = :645 an Φ(Z ff= ) = ff proportion of t Λ n;b values smaller than t n is roughly a measure of meian bias of t n in normal units." When exactly half of the bootstrap samples have t Λ n;b values less than t n, then Z 0 = 0.! a = Pn j= (t n;( ) t n; j ) P n j= (t n;( ) t n; j ) 3 5 3= is the estimate acceleration, where t n; j an is the value of t n when the j-th case is remove from the sample t n;( ) = n Pn j= (t n; j ) 7 8

13 ffl BC a intervals are secon orer accurate P rf < lower en of BC a intervalg = ff + C lower n P rf > upper en of BC a intervalg = ff + C upper n ffl Bootstrap percentile intervals are first orer accurate P rf < lower eng ff = + CΛ lower p n ffl ABC intervals are approximations to BC a intervals secon orer accurate only use about 3% of the computation time (See Efron & Tibshirani (993) Chapter 4) P rf < upper eng ff = + CΛ upper p n 9 30 # This is S-plus coe for creating # bootstrappe confience intervals # for a correlation coefficient. It is # store in the file # # lawschl.ssc # # Any line precee with a poun sign # is a comment that is ignore by the # program. The law school ata are # rea from the file # # lawschl.at # Enter the law school ata into a ata frame laws <- rea.table("lawschl.at", col.names=c("school","lsat","gpa")) laws School LSAT GPA

14 # Plot the ata par(fin=c(7.0,7.0),pch=6,mkh=.5,mex=.5) plot(laws$lsat,laws$gpa, type="p", xlab="gpa",ylab="lsat score", main="law School Data") # Compute the sample correlation matrix rr<-cor(laws$lsat,laws$gpa) cat("estimate correlation: ", roun(rr,5), fill=t) # First test for zero correlation n <- length(laws$lsat); tt<- sqrt(n-)*rr/sqrt( - rr*rr) pval <- - pt(tt,n-) pval <- roun(pval,igits=5) cat("t-test for zero correlation: ", roun(tt,4), fill=t) t-test for zero correlation: cat("p-values for the t-test for zero correlation: ", pval, fill=t) Estimate correlation: p-values for the t-test for zero correlation: # Use Fisher's z-transformation to construct # approximate confience intervals. # First set the level of confience at -alpha. alpha <-.0 z <- 0.5*log((+rr)/(-rr)) zl <- z - qnorm(-alpha/)/sqrt(n-3) zu <- z + qnorm(-alpha/)/sqrt(n-3) rl <- roun((exp(*zl)-)/(exp(*zl)+), igits=4) ru <- roun((exp(*zu)-)/(exp(*zu)+), igits=4) per <- (-alpha)*00; cat( per,"% confience interval: (", rl,", ",ru,")",fill=t) # Compute bootstrap confience intervals. # Use B=5000 bootstrap samples. nboot < rboot <- bootstrap(ata=laws, statistic=cor(gpa,lsat),b=nboot) Forming replications to 00 Forming replications 0 to 00 Forming replications 0 to Forming replications 480 to 4900 Forming replications 490 to % confience interval: ( 0.509, ) 35 36

15 # limits.emp(): Calculates empirical percentiles # for the bootstrappe parameter # estimates in a resamp object. # The quantile function is use to # calculate the empirical percentiles. # usage: # limits.emp(x, probs=c(0.05, 0.05, 0.95, 0.975)) limits.emp(rboot, probs=c(0.05,0.95)) 5% 95% Param # limits.bca(): Calculates BCa (bootstrap # bias-correct, ajuste) # confience limits. # usage: # limits.bca(boot.obj, # probs=c(0.05, 0.05, 0.95, 0.975), # etails=f, z0=null, # acceleration=null, # group.size=null, # frame.eval.jack=sys.parent()) # Do another set of 5000 bootstrappe values limits.bca(rboot,probs=c(0.05,0.95),etail=t) rboot <- bootstrap(ata=laws, statistic=cor(gpa,lsat),b=nboot) limits.emp(rboot, probs=c(0.05,0.95)) $limits: 5% 95% Param % 95% Param $emp.probs: 5% 95% Param # Both sets of confience intervals coul # have been obtaine from the summary( ) # function summary(rboot) $z0: Param $acceleration: Param $group.size: [] Call: bootstrap(ata = laws, statistic = cor(gpa, LSAT), B = nboot) Number of Replications: 5000 Summary Statistics: Observe Bias Mean SE Param Empirical Percentiles:.5% 5% 95% 97.5% Param BCa Percentiles:.5% 5% 95% 97.5% Param

16 5000 Bootstrap Correlations # Make a histogram of the bootstrappe # correlations. hist(rboot$rep,nclass=50, xlab=" ", main="5000 Bootstrap Correlations", ensity=.000) The bootstrap can fail: Example: X ; X ; : : : ; X n are sample from a uniform (0; ) istribution: true ensity: f(x) = ; 0 < x < true c..f.: F (x) = 8 >< >: 0 x» 0 x 0 < x» x > Bootstrap percentile confience intervals for ten to be too short. Application of the bootstrap must aequately replicate the ranom process that prouce the original sample ffl Simple ranom samples ffl Neste" experiments Sample plants from a fiel Sample leaves from plants ffl Curve fitting (existence of covariates) Fixe levels Ranom samples 43 44

17 Parametric Bootstrap ffl Suppose you knew" that (X j ; X j ) j = ; : : : ; n were obtaine from a simple ranom sample from a bivariate normal istribution, i.e., X j = 6 4 X j X j ο NID μ μ ; ± ffl Estimate unknown parameters 3 μ μ = 6 n X 4 7 μ X j = X j= ± = n n X 5 = n C A j= (X j X)(X j X) T 45 ffl Obtain a bootstrap sample of size n by sampling from a N(^μ; ^±) istribution, i.e. X ;b ; : : : ; X n;b an compute t Λ n;b = r n;b ffl Repeat this to obtain r n; ; r n; ; : : : ; r n;b ffl Compute bootstrap stanar errors bias estimators confience intervals 46 References Davison, A.C. an Hinkley, D.V. (997) Bootstrap Methos an Their Applications, Cambrige Series in Statistical an Probabilistic Mathematics, Cambrige University Press, New York. Efron, B. (98) The Jackknife, The Bootstrap an other resampling plans, CBMS, 38, SIAM-NSF, Philaelphia. Efron, B. an Gong, G. (983) The American Statistician,, Efron, B. (987) Better Bootstrap confience intervals (with iscussion) Journal of the American Statistical Association, 8, Efron, B. an Tibsharani, R. (993) An Introuction to the Bootstrap, Chapman an Hall, New York. Shao, J. an Tu, D. (995) The Jackknife an Bootstrap, New York, Springer. 47 Example.: Stormer viscometer ata (Venables & Ripley, Chapter 8) ffl measure viscosity of a flui ffl measure time taken for an inner cycliner in the mechanism to complete a specific number of revolutions in response to an actuating weight ffl calibrate the viscometer using runs with varying weights (W ) (g) fluis with known viscosity (V ) recor the time (T ) (sec 48

18 ffl theoretical moel T = fi V W fi + ffl # This coe is use to explore the # Stormer viscometer ata. It is store # in the file # # stormer.ssc # # Enter the ata into a ata frame. # The ata are store in the file # # stormer.at library(mass) stormer <- rea.table("stormer.at") stormer 49 Viscosity Wt Time Starting values for fi an fi : Fit an approximate linear moel. Note that T fi V i i = + ffl W i fi i ) (W i fi )T i = fi V i + ffl i (W i fi ) ) W i T i = fi V i + fi T i + ffl i (W fi ) " " this is the new this is the response variable new error # Use a linear approximation to obtain # starting values for least squares # estimation in the non-linear moel fm0 <- lm(wt*time ~ Viscosity + Time -, ata=stormer) b0 <- coef(fm0) names(b0) <- c("b","b") # Fit the non-linear moel storm.fm <- nls( formula = Time ~ b*viscosity/(wt-b), ata = stormer, start = b0, trace = T) Use OLS estimation to obtain fi (0) = 8:876 fi (0) = : : : :

19 # Create a bivariate confience region # for the the (b,b) parameters. # First set up a gri of (b,b) values bc <- coef(storm.fm) se <- sqrt(iag(vcov(storm.fm))) v <- eviance(storm.fm) summary(storm.fm)$parameters Value St. Error t value b b gsize<-5 b <- bc[] + seq(-3*se[], 3*se[], length = gsize) b <- bc[] + seq(-3*se[], 3*se[], length = gsize) bv <- expan.gri(b, b) # Create a function to evaluate sums of squares ssq <- function(b) sum((stormer$time - b[] * stormer$viscosity/ (stormer$wt-b[]))^) # Create the plot # Evalute the sum of square resiuals an # approximate F-ratios for all of the # (b,b) values on the gri beta <- apply(bv,, ssq) n<-length(stormer$viscosity) f<-length(bc) f<-n-f fstat <- matrix( ((beta - v)/f) / (v/f), gsize, gsize) par(fin=c(7.0,7.0), mex=.5,lw=3) plot(b, b, type="n", main="95% Confience Region") contour(b, b, fstat, levels=c(,,5,7,0,5,0), labex=0.75, lty=, a=t) contour(b, b, fstat, levels=qf(0.95,,), labex=0, lty=, a=t) text(3.6,0.3,"95% CR", aj=0, cex=0.75) points(bc[], bc[], pch=3, mkh=0.5) # remove b,b, an bv rm(b,b,bv,fstat) 55 56

20 Construct a joint confience region for (fi ; fi ): ffl Deviance (resiual sum of squares): (fi ; fi ) = n X 6 j= T 4 j fi V j W j fi ffl Approximate F-statistic: F (fi ; fi ) = (fi;fi) ( c fi; c fi) ( c fi; c fi) n b % Confience Region % CR ffl An approximate ( ff) 00% confience interval consists of all (fi ; fi ) such that b F (fi ; fi ) < F (;n );ff Bootstrap Estimation Bootstrap I: Sample n = 3 cases (T i ; W i ; V i ) from the original ata set (using simple ranom sampling with replacement) #========================================= # Bootstrap I: # Treat the regressors as ranom an # resample the cases (y,x_,x_) #========================================= storm.boot <- bootstrap(stormer, coef(nls(time~b*viscosity/(wt-b), ata=stormer,start=bc)),b=000) summary(storm.boot) Call: bootstrap(ata = stormer, statistic = coef(nls(time ~ (b * Viscosity)/(Wt - b), ata = stormer, start = bc)), B = 000) Number of Replications:

21 Summary Statistics: Observe Bias Mean SE b b Empirical Percentiles:.5% 5% 95% 97.5% b b BCa Percentiles:.5% 5% 95% 97.5% b b Correlation of Replicates: b b b b # Prouce histograms of the bootstrappe # values of the regression coefficients # The following coe is use to raw # non-parametric estimate ensities, # normal ensities, an a histogram # on the same graph. # truehist(): Plot a Histogram (prob=t # by efault) For the function # hist(), probability=f by # efault. # with.sj(): Banwith Selection by Pilot # Estimation of Derivatives. # Uses the metho of Sheather # & Jones (99) to select the # banwith of a Gaussian kernel # ensity estimator. 6 6 b.boot <- storm.boot$rep[,] b.boot <- storm.boot$rep[,] library(mass) par(fin=c(7.0,7.0),mex=.5) # ensity() : Estimate Probability Density # Function. Returns x an y # coorinates of a non-parametric # estimate of the probability # ensity of the ata. Options # inclue the type of winow to # use an the number of points # at which to estimate the ensity. # n = the number of equally space # points at which the ensity # is evaluate. mm <- range(b.boot) min.int <- floor(mm[]) max.int <- ceiling(mm[]) truehist(b.boot,xlim=c(min.int,max.int), ensity=.000) with.sj(b.boot) [] b.boot.ns <- ensity(b.boot,n=00,with=.6) b.boot.ns <- list( x = b.boot.ns$x, y = norm(b.boot.ns$x,mean(b.boot), sqrt(var(b.boot)))) 63 64

22 # Draw the non-parametric ensity lines(b.boot.ns,lty=3,lw=) # Draw normal ensities lines(b.boot.ns,lty=,lw=) truehist(b.boot,xlim=c(min.int,max.int), ensity=.00) with.sj(b.boot) [] legen(7.,-0.30,c("nonparametric", "Normal approx."), lty=c(3,),bty="n",lw=) # Display the istribution of the estimate # of the secon parameter par(fin=c(7.0,7.0),mex=.5) mm <- range(b.boot) min.int <- floor(mm[]) max.int <- ceiling(mm[]) b.boot.ns <- ensity(b.boot,n=00,with=.8) b.boot.ns <- list( x = b.boot.ns$x, y = norm(b.boot.ns$x,mean(b.boot), sqrt(var(b.boot)))) lines(b.boot.ns,lty=3,lw=) lines(b.boot.ns,lty=,lw=) legen(.,-0.30,c("nonparametric", "Normal approx."), lty=c(3,),bty="n",lw=) b.boot b.boot Nonparametric Normal approx. Nonparametric Normal approx

23 Bootstrap II: Fix the values of the explanatory variables f(w j ; V j ) : j = ; : : : ; ng ffl Compute resiuals from fitting the moel to the original sample e j = T j fi V j W j fi j = ; : : : ; n ffl Approximate sampling from the population of ranom errors by taking a sample (with replacement) from fe ; : : : ; e n g say, ffl Create new observations: where (T Λ j;b ; W j; V j ) j = ; : : : ; n T Λ j;b = fi V j W j fi + e Λ j ffl Fit the moel to the j-th bootstrap sample to obtain fi Λ ;b an fi Λ ;b ffl Repeat this for b = ; : : : ; B bootstrap samples e Λ ;b ; eλ ;b ; : : : ; eλ n;b #======================================= # Bootstrap II: # Treat the regressors as fixe an # resample from the resiuals #======================================= # Center the resiuals at zero an # ivie resiuals by linear approximations # to a multiple of the stanar errors. These # centere an scale resiuals approximately # have the same first two moments as the # ranom errors, but they are not quite # uncorrelate. rs <- scale(resi(storm.fm),center=t,scale=f) g <- gra.f(b[], b[], stormer$viscosity, stormer$wt, stormer$tim) D <- attr(g, "graient") h <- -iag(d%*%solve(t(d)%*%d)%*%t(d)) rs <- rs/sqrt(h) fe <- length(rs)-length(coef(storm.fm)) vr <- var(rs) rs <- rs%*%sqrt(eviance(storm.fm)/fe/vr) gra.f <- eriv3( expr = Y ~ (b*x)/(x-b), namevec = c("b", "b"), function.arg = function(b, b, X, X, Y) NULL) 7 7

24 # Compute 000 bootstrappe values of the # regression parameters # Create a function to use in fitting # the moel to bootstrap samples storm.boot <- bootstrap(ata=rs, statistic=storm.bf(rs),b=000) storm.bf <- function(rs) assign("tim", fitte(storm.fm) + rs, frame = ) nls(formula = Tim ~ (b*viscosity)/(wt-b), ata = stormer, start = coef(storm.fm) )$parameters } summary(storm.fm)$parameters Forming replications to 00 Forming replications 0 to 00 Forming replications 0 to 300 Forming replications 30 to 400 Forming replications 40 to 500 Forming replications 50 to 600 Forming replications 60 to 700 Forming replications 70 to 800 Forming replications 80 to 900 Forming replications 90 to 000 Value St. Error t value b b Call: bootstrap(ata = rs, statistic = storm.bf(rs), B = 000) 73 b.boot <- storm.boot$rep[,] b.boot <- storm.boot$rep[,] 74 # The BCA intervals may not be correctly # compute by the following function summary(storm.boot) Number of Replications: 000 Summary Statistics: Observe Bias Mean SE b b Empirical Percentiles:.5% 5% 95% 97.5% b b BCa Confience Limits:.5% 5% 95% 97.5% b b Correlation of Replicates: b b b

25 # The following coe is use to raw # non-parametric estimate ensities, # normal ensities, an histograms # on the same graphic winow. par(fin=c(7.0,7.0),mex=.3) # Draw a non-parametric ensity mm <- range(b.boot) min.int <- floor(mm[]) max.int <- ceiling(mm[]) lines(b.boot.ns,lty=3,lw=) # raw normal ensities truehist(b.boot,xlim=c(min.int,max.int), ensity=.000) b.boot.ns <- ensity(b.boot,n=00, with=with.sj(b.boot)) b.boot.ns <- list( x = b.boot.ns$x, y = norm(b.boot.ns$x,mean(b.boot), sqrt(var(b.boot)))) lines(b.boot.ns,lty=,lw=) legen(7,-0.,c("nonparametric", "Normal approx."), lty=c(3,),bty="n",lw=) # Display the istribution for the other # parameter estimate par(fin=c(7.0,7.0),mex=.3) mm <- range(b.boot) min.int <- floor(mm[]) max.int <- ceiling(mm[]) truehist(b.boot,xlim=c(min.int,max.int), ensity=.0000) b.boot.ns <- ensity(b.boot,n=00, with=with.sj(b.boot)) b.boot.ns <- list( x = b.boot.ns$x, y = norm(b.boot.ns$x,mean(b.boot), sqrt(var(b.boot)))) lines(b.boot.ns,lty=3,lw=) lines(b.boot.ns,lty=lw=) legen(0.5,-0.5,c("nonparametric", "Normal approx."),lty=c(3,), bty="n",lw=) b.boot Nonparametric Normal approx. 80

26 Comparison of stanar errors: Asymptotic Para- normal Ranom Fixe meter approx. Bootstrap Bootstrap fi fi b.boot Nonparametric Normal approx. Comparison of approximate 95% confience intervals: Asymptotic Para- normal Ranom Fixe meter approx. Bootstrap Bootstrap fi (7.50, 3.3) (8.3, 30.40) (7.4, 3.03) fi (0.83, 3.60) (.06, 3.5) (.03, 3.46) " fi i ± t ;:05 S c fii 8 8