Point and Interval Estimation II - RevisedBios 662Michael G. Hudgens, [email protected]://www.bios.unc.edu/∼mhudgens2006-09-20 11:01BIOS 662 1 Point and Interval Estimation II - RevisedOutline• Nonparametric CIs for median– Exact– Large sample• CIs for varianceBIOS 662 2 Point and Interval Estimation II - RevisedNonparametric CI for the Median• Suppose X1, . . . , Xniid according to continuous distri-bution F• Let ζ1/2be the population median• We will showPr[X(r)< ζ1/2< X(n−r+1)] =12nn−rXi=rni• Therefore, for fixed n, we choose largest r such that12nn−rXi=rni≥ 1 − αBIOS 662 3 Point and Interval Estimation II - RevisedBernoulli RV• Let Y be a Bernoulli r.v.• Y can take on two values, 0 or 1Pr[Y = 1] = π; Pr[Y = 0] = 1 − πE(Y ) = π; V ar(Y ) = π(1 − π)BIOS 662 4 Point and Interval Estimation II - RevisedBinomial RV• Process that produces independent Bernoulli RVs withthe same probability of success π• Let Y count the number of successes in n trials• Y ∼ Binomial(n,π)Pr[Y = y] =nyπy(1 − π)n−yE(Y ) = nπ; V ar(y) = nπ(1 − π)BIOS 662 5 Point and Interval Estimation II - RevisedDerivation of CI for Median• CDFPr[Xi≤ x] = F (x)• ThereforePr[X(r)≤ x] = Pr[at least r of the Xi≤ x]=Pni=rniF (x)i{1 − F (x)}n−iBIOS 662 6 Point and Interval Estimation II - RevisedDerivation of CI for Median• By law of total probabilityPr[X(r)≤ ζp] = Pr[X(r)≤ ζp, X(s)≥ ζp]+ Pr[X(r)≤ ζp, X(s)< ζp]• If s > r, then X(s)< ζp⇒ X(r)< ζp• ThereforePr[X(r)≤ ζp] = Pr[X(r)≤ ζp≤ X(s)]+ Pr[X(s)< ζp]BIOS 662 7 Point and Interval Estimation II - RevisedDerivation of CI for MedianPr[X(r)≤ ζp≤ X(s)] = Pr[X(r)≤ ζp] − Pr[X(s)< ζp]=Pni=rniF (ζp)i{1 − F (ζp)}n−i−Pni=sniF (ζp)i{1 − F (ζp)}n−i=Ps−1i=rniF (ζp)i{1 − F (ζp)}n−i• If p = 1/2; F (ζp) = 1/2, such thatPr[X(r)≤ ζ0.5≤ X(s)] =12ns−1Xi=rniBIOS 662 8 Point and Interval Estimation II - RevisedDerivation of CI for Median• We could choose any r and s such thatPr(X(r)≤ ζ0.5≤ X(s)) =12ns−1Xi=rni≥ 1 − α• But the best choice for s is n − r + 1 (why?)• Thus we choose r such that12nn−rXi=rni≥ 1 − αBIOS 662 9 Point and Interval Estimation II - RevisedDerivation of CI for Median• Values of r for 95% CI for Mediann r1-5 06-8 19-11 212-14 315-16 417-19 520-22 623-24 725-27 828-29 930-32 1033-34 11• Cf page 269-270 van Belle et al.BIOS 662 10 Point and Interval Estimation II - Revised95% CI for Betacarotene Example• For n = 23, choose r = 7 such that n − r + 1 = 17• Therefore(y(7)= 106, y(17)= 186)gives a 95% CI for the median betacarotene value• This CI makes no assumptions about the distributionof the Y ’s• Note:122323−7Xi=723i= 0.9653 ≥ 1 − α> sum(dbinom(7:16,23,1/2))[1] 0.9653103BIOS 662 11 Point and Interval Estimation II - RevisedSAS Code and Outputproc univariate data=beta cipctldf;var base1;run;95% Confidence Limits -------Order Statistics-------Quantile Distribution Free LCL Rank UCL Rank Coverage99% . . . . .95% 212 298 21 23 58.7590% 202 298 19 23 83.8375% Q3 162 252 13 22 97.3550% Median 106 186 7 17 96.5325% Q1 74 124 2 11 97.3510% 68 92 1 5 83.835% 68 80 1 3 58.751% . . . . .0% MinBIOS 662 12 Point and Interval Estimation II - RevisedLarge sample CI for median• The above method of finding a (1 − α)100% CI for themedian is exact, i.e., the probability the CI contains ζ.5is guaranteed to be at least (1 − α)• Now we derive a large sample CI for the median usingthe CLT• This will be approx in that the probability the CI con-tains ζ.5is ≈ (1 − α), with the approx improving asn → ∞BIOS 662 13 Point and Interval Estimation II - RevisedLarge sample CI for any quantile• If general,Pr[ζp< Z(r)] =Pr−1i=0niF (ζp)i{1 − F (ζp)}n−i=Pr−1i=0nipiqn−iwhere q = 1 − p• From CLT, if Y ∼ Bin(n, p), thenY − np + 1/2√npq∼ N(0, 1)• ThusPr[ζp≤ Z(r)] = Pr[Y ≤ r − 1]= Pr[Z ≤(r−1)−np+1/2√npq]= Φ(r−np−1/2√npq)BIOS 662 14 Point and Interval Estimation II - RevisedLarge sample CI for any quantile• Goal is symmetric (1 − α)% CI, so wantα/2 = Pr[ζp< Z(r)] = Φ(r − np − 1/2√npq)• That is−z1−α/2=r − np − 1/2√npq• Implyingr = np +12− z1−α/2√npq• For p = 1/2, yieldsr =n + 12− z1−α/2√n2BIOS 662 15 Point and Interval Estimation II - RevisedLarge sample CI for any quantile• Similar reasoning yieldss = np +12+ z1−α/2√npq• Thus (1 − α)% CI for ζpis given by(X(brc), X(dse))• Note n large enough ensures brc, dse ∈ {1, . . . , n}BIOS 662 16 Point and Interval Estimation II - RevisedLarge Sample CI for Median: Example• Suppose n = 100 and α = 0.05• Thenz1−α/2√n2= 5(1.96) = 9.8• Rounding yields:50.5 ± 9.8 ⇒ (y(40), y(61))• Can show r = 40 using exact method> sum(dbinom(40:60,100,1/2))[1] 0.9647998> sum(dbinom(41:59,100,1/2))[1] 0.943112BIOS 662 17 Point and Interval Estimation II - RevisedCI for Variance• Recall (result 4.4 p.95 text)(n − 1)s2σ2∼ χ2n−1• Therefore1 − α = Pr[χ2α/2,n−1≤(n − 1)s2σ2≤ χ21−α/2,n−1]• Implying1 − α = Pr(n − 1)s2χ21−α/2,n−1≤ σ2≤(n − 1)s2χ2α/2,n−1BIOS 662 18 Point and Interval Estimation II - RevisedCI for Variance• Since the χ2distribution is not symmetric, need to lookup both χ2α/2,n−1and χ21−α/2,n−1• This CI is dependent on the Y ’s being from a normaldistributionBIOS 662 19 Point and Interval Estimation II - RevisedCI for Variance for Betacarotene Example• n = 23; s2= 3701.36• χ2.025,22= 10.98; χ2.975,22= 36.78• Therefore, 95% CI for σ2(22(3701.36)/36.78, 22(3701.36)/10.98) = (2213.973, 7416.203)• 95% CI for σ = (47.05, 86.12)BIOS 662 20 Point and Interval Estimation II - RevisedSAS Code and Outputproc univariate data=beta cibasic;var base1;run;Basic Confidence Limits Assuming NormalityParameter Estimate 95% Confidence LimitsMean 150.78261 124.47394 177.09128Std Deviation 60.83880 47.05242 86.10828Variance 3701 2214 7415BIOS 662 21 Point and Interval Estimation II - RevisedCI for Variance - Nonnormal data• Large sample theory√n(s2n− σ2) →dN(0, (α4− 1)σ4)where α4= E(X−µ)4/σ4is the kurtosis (cf. Dudewiczand Mishra Modern Mathematical Statistics, p. 325)• “Crude approximation”: replace usual CI with(n − 1)s2χ21−α/2,n−1(1 +