Point and Interval Estimation IBios 662Michael G. Hudgens, [email protected]://www.bios.unc.edu/∼mhudgens2007-08-24 16:46BIOS 662 1 Point and Interval Estimation IInference• Inference: Using statistics and probability theory todraw conclusions about parameters• Two modes of inference:– Estimation: attempt to es timate value of parame-ter(s) and quantify uncertainty about these estimate(s)– Hypothesis testing: posit certain values for parame-ters and test whether the observed data are consistentwith the hypothesisBIOS 662 2 Point and Interval Estimation IEstimation• Estimand: parameter of interest we are trying to esti-mate; a constant; Eg µ• Estimator: the statistic used to estimate the estimand;a random variable; Eg¯Y• Estimate: a realization of an estimator from an ob-served data set ; Eg¯y = 36.3BIOS 662 3 Point and Interval Estimation IEstimating µ• Suppose Y1, . . . , Ynare a random sample from a distri-bution with mean µ• The estimator¯Y is an unbiased estimator of µ, i.e.,E(¯Y ) = µi.e., the mean of the sampling distribution of¯Y equalsµ, the population parameter of interestBIOS 662 4 Point and Interval Estimation IConfidence Interval for µ• Suppose Y1, . . . , Ynare a random sample from a normaldistribution with mean µ and variance σ2• Then¯Y ∼ N(µ,σ2n)cf corollary from last set of slides• We can use t his to derive a confidence interval (CI) forµBIOS 662 5 Point and Interval Estimation IConfidence Interval for µ• First define zpsuch thatPr[Z ≤ zp] = pfor Z ∼ N(0, 1); by symmetry, zp= −z1−p• zpis the pthquantile of a standard normal distributionBIOS 662 6 Point and Interval Estimation IConfidence interval for µ1 − α = Pr[−z1−α/2< Z < z1−α/2]= Pr[−z1−α/2<¯Y −µσ/√n< z1−α/2]= Pr[−z1−α/2σ√n<¯Y − µ < z1−α/2σ√n]= Pr[−¯Y − z1−α/2σ√n< −µ < −¯Y + z1−α/2σ√n]= Pr[¯Y − z1−α/2σ√n< µ <¯Y + z1−α/2σ√n]BIOS 662 7 Point and Interval Estimation IConfidence interval for µ• 100(1 − α)% CI for µ¯Y ± z1−α/2σ√nor(¯Y − z1−α/2σ√n,¯Y + z1−α/2σ√n)• Values of z1−α/2α z1−α/20.10 1.6450.05 1.9600.01 2.576BIOS 662 8 Point and Interval Estimation ICI Interpretation, Comment• van Belle et al (p 86): The probability is 1 −α that theinterval straddles the population mean µ• If we draw 100 different random samples, on average100(1 − α)% of them will contain µ• To decrease the width of CI:– increase α, i.e., de crease confidence– increase sample sizeBIOS 662 9 Point and Interval Estimation ICI Example• Example 4.8 text: SIDS birthweights• n = 78,¯Y = 2994g, σ = 800g• A 95% CI for the mean birthweight2994 ± 1.96800√78= (2816, 3172)• A 99% CI for the mean birthweight2994 ± 2.58800√78= (2760, 3228)BIOS 662 10 Point and Interval Estimation IAssumptions• Y ’s are sampled from a normal distribution• Variance is known• What do we do if variance is unknown?• If σ2is not known, we can estimate it with s2• However, the distribution of¯Y − µs/√nis not the normal distribution!BIOS 662 11 Point and Interval Estimation IDistribution of s2• Result 4.4 (p 95 text): If a random variable Y is nor-mally distributed with mean µ and variance σ2, thenfor a random sample of size n, the quantity(n − 1)s2σ2has a chi-square distribution with n −1 degrees of free-dom, which we denote by χ2n−1BIOS 662 12 Point and Interval Estimation Iχ2Distribution0 2 4 6 8 10 12 140.0 0.1 0.2 0.3 0.4 0.5Xdensitydf=1df=2 (bold)df=4df=10BIOS 662 13 Point and Interval Estimation It Distribution• Let Z ∼ N(0, 1) and W ∼ χ2ν• If Z and W are independent, thenT =ZpW/νwill follow the t-distribution with ν degrees of freedom.• We knowZ =¯Y − µσ/√n∼ N(0, 1) and W =(n − 1)s2σ2∼ χ2n−1• Can show¯Y and s2are independentBIOS 662 14 Point and Interval Estimation It Distribution−3 −2 −1 0 1 2 30.0 0.1 0.2 0.3 0.4 0.5zdensityN(0,1)t(df=2)BIOS 662 15 Point and Interval Estimation ICI for µ when σ2unknown• Substituting, we getT =ZpW/ν=√n(¯Y − µ)/σp{(n − 1)s2/σ2}/(n − 1)=¯Y − µs/√n• Thus (1 − α ) × 100% CI is given by¯Y ± tn−1,1−α/2s√n• Note 1: We are still assuming the Y ’s are normal• Note 2: If n ≥ 30, can use z instead of tBIOS 662 16 Point and Interval Estimation IExample•¯Y = 150.8, s = 60.84, n = 23• t22,.975= 2.07• 95% CI for µ:150.8 ± 2.0760.84√23= 150.8 ± 26.3 = (124.5, 177.1)• Note here we multiply (estimated) s.e. by 2.07 ratherthan the usual 1.96 as a penalty for not knowing σBIOS 662 17 Point and Interval Estimation INon-normal data• If the Y ’s are not normally distributed, we use the CLT:• If Y1, . . . , Ynis a random sample from a distributionwith E(Yi) = µ and V ar(Yi) = σ2for i = 1, . . . , n,then¯Y is approximately distributed as N(µ, σ2/n) forlarge n• The use of the CLT to construct a CI for µ requiresknowledge of σ2• To use the CLT when σ2unknown requires Slutsky’sTheoremBIOS 662 18 Point and Interval Estimation ISlutsky’s Theorem• If Xnis a sequence of r.v. that converges in distributionto F (x) and• Ynis a sequence of r.v. that converges in probability toa constant c,• then Wn= XnYnconverges in distribution to F (cw)• I.e.limn→∞Pr[Wn≤ w] = F (cw) = limn→∞Pr[Xn≤ cw]BIOS 662 19 Point and Interval Estimation INon-normal Data• LetXn=¯Y − µσ/√nand Yn=sσ2s2• Know Xn→dZ ∼ N(0, 1) and σ2/s2→p1• Then Slutsky’s Theorem impliesWn= XnYn=¯Y − µσ/√nsσ2s2=¯Y − µs/√nwill be approximately ∼ N(0, 1)• The approximation gets better as n → ∞BIOS 662 20 Point and Interval Estimation ILarge Sample CI for µ• If n is sufficiently large, an approximate 100(1 − α)%CI for µ is¯Y ± z1−α/2s√n• This is true regardless of the original distribution of theY ’sBIOS 662 21 Point and Interval Estimation IExample• A survey was conducted to estimate the mean age thatsmoking was started among women who smoke . A ran-dom sample of 243 smoking women in NC found¯y =16.8 and s = 2.36.• A 95% CI for the mean age of smoking onset is:16.8 ± 1.962.36√243= (16.5, 17.1)BIOS 662 22 Point and Interval Estimation ISummaryNormal σ2known n large Confidence Interval√ √¯Y ± z1−α/2(σ/√n)√¯Y ± tn−1,1−α/2(s/√n)√ √¯Y ± z1−α/2(σ/√n)√¯Y ± z1−α/2(s/√n)Transform, nonparametricsBIOS 662