5. 추정 (Estimation) 2014/4/17

Transcription

1 5. 추정 (Estimatio) 014/4/17

2 5.1 머리말 (Itoriductio) 통계적추측 (statistical iferece) 어느모집단으로부터구한표본에서얻어진결과를기초로그모집단에관해추측하는과정 Say somethig about the populatio based o the iformatio of the sample 1) 추정 (estimatio) ) 가설검정 (hypothesis testig)

3 추정치 (estimate) 1) 점추정 (poit estimate) ) 구간추정 (iterval estimate) 추정식 (estimator) 불편이성 (ubiasedess) x 3 Target populatio vs. samplig populatio x i 의추정식 ˆ( rv.. based o data) is a ubaised estimator of (parameter) if E( ˆ ) ex. E( X ), so sample mea is a ue of the populatio mea if the samples are radomly selected from N(, )

4 예 ) sample variace ( of So Es ( ) Bias = 1 E ( yi y) E( ˆ ) s ) is a ubiased estimator : ot a ubiased estimator Bias of a ubiased estimator is zero Probability samplig ad o-probability samplig Radomizatio Blidig

5 5. 모집단평균의신뢰구간 (Cofidece iterval of populatio mea) 신뢰구간 = 추정평균 ( 신뢰도계수) 표준오차 cof. it.=eatimated value (reliability coef) SE x z (1 ) x If we select samples repeatedly from ormal populatio, x z(1 ) x will iclude with the probability of 100(1- )% 1 :cofidece level (ex..95) 신뢰수준 :sigificace level (ex..05) 유의수준

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7 < 보기 5..1> A researcher measures amout of a certai ezyme. =10, sample mea=, We ca assume ormality with pop variace= % C.I. of? 45 x x (17.76, 6.4) 100

8 < 보기 5..> Measurig maximum stregth of a certai muscle. We wat 99% CI of the pop mea. We assume ormality with pop variace=144. =15, sample mea=84.3, z=.58 with 0.99 cofidece level, SE= x What is the 99% C.I. of? (3.10) (76.3, 9.3)

9 Sample from o-ormal pop cetral limit theorem < 보기 5..3> delay time because of patiet s beig late at a cliic, =35, sample mea=17. mi, sd from the previous study (assumed to be kow)=8 mi. Pop is ot ormally dist ed. what is 90% CI of? 8/ x (1.35) , 19.4

10 < 보기 5..4> Measure activity of a certai ezyme from 35 patiets Populatio variace= 0.36, 95% CI? -> apply CLT (.6/ 35) (0.5174, )

11 CI calculatio usig R > m < > s <- sqrt(0.36) > <- 35 > alpha < > error <- qorm(1-alpha/)*s/sqrt() > left <- m-error > right <- m+error > left [1] > right [1] > cofit <- fuctio(m,s,,alpha=0.05){ error <- qorm(1-alpha/)*s/sqrt() left <- m-error right <- m+error prit(c(left,right)) } cofit(0.7164,sqrt(0.36),35) [1]

12 5.3 t- 분포 (t-dist ) Pop variace is kow ad is large: Pop variace is ot kow ad is large : Small sample size (<30) : derived by Gosset Studet s t-dist 표 E ( ) 1 i x x s 1 x t t s x z

13 Some properties of t-dist 1) mea= 0 ) Symmetric about the mea 3) Variace > 1, -> 1 as -> 4) t 5) The shape depeds o degrees-offreedom)=-1 x <- seq(-4, 4, legth=100) hx <- dorm(x) degf <- c(1, 3, 8, 30) colors <- c("red", "blue", "darkgree", "gold", "black") labels <- c("df=1", "df=3", "df=8", "df=30", "ormal") plot(x, hx, type="l", lty=, xlab="x value", ylab="desity", mai="compariso of t Distributios") for (i i 1:4){ lies(x, dt(x,degf[i]), lwd=, col=colors[i]) } leged("topright", iset=.05, title="distributios", labels, lwd=, lty=c(1, 1, 1, 1, ), col=colors)

14 6) Flatter at the ceter ad heavier tails tha ormal dist 7) t-dist -> ormal dist as -1 -> CI : x t (1 ) s

15 < 보기 5.3.1> =15, measure Amylase, sample mea=96uit/100ml, sd= 35, pop variace is ot kow. 95% CI of the pop mea? SE( x)= s df= 1 14 t (9.04) (77,115)

16 Choice of z ad t pop~ormal? eough? eough? σ kow? σ kow? σ kow? σ kow? applyig CLT No-parametric methods

17 5.4 CI of the differece of the two meas Samples from ormal pop s 1 x1 x z1 1 < 보기 5.4.1> Measure serum uric acid from 1 patiets x1 4.5 ml /100ml, measuremets from 15 ormal cotrols x 3.4, variaces are kow to be 1 for each group, 95% CI for 1-? ( ) (.39) (.3, 1.9) 1 15 CI does ot iclude 0

18 Sample from o-ormal pop cetral limit theorem < 보기 5.4.> To compare socio-ecoomic status (SES) of patiets from two hospitals. 75 pts from hospital A: x1 6,800, 80 pts from hospital B: x 4, 450, pop variaces are 1 99% CI of -? 1 600, 500, (600) (500) ( ).58 (10, 580) 75 80

19 t-dist ad differece of the meas: I practice, pop variace is typically ukow. Two approaches 1) Same variaces, ) Differet variaces 1) Whe the variaces are the same: we calculate pooled estimate by calculatig weighted average of the variaces s ( 1 1) s1 ( 1) s p 1

20 100(1 )% CI of 1 sp sp ( x1 x ) t(1 ) df= 1 1 < 보기 5.4.3> to measure Amylase: 15 ormal cotrols(group) sample mea=96, sd=35. pts(group1) sample mea ad sd=10 ad 40. pop ormal. Variaces ukow but equal. 100(1 )% CI of? s p 1 14(35) 1(40) (10 96) ,50 15

21 ) Whe the variaces are differet ( x x ) ( ) s s 1 does ot follow t-dist! w1 t1 wt t ' 1 w1 w ( x1 x ) t '(1 ) s 1 s * w s, w s, df 1 t t, df 1 t t

22 < 보기 5.4.3> To measure some biomarker. We ca assume ormal dist. But variaces are ot the same. 95% CI of? 1 Pts ormal t 9 t (.6) (.0930) t ' ( ).55 ( 9., 40.0) 10 0

23 pop~ormal? σ kow? applyig CLT No-parametric methods

24 숙제

25 5.5 모집단비율의신뢰구간 (CI of proportio) 100(1 -)% CI of p? p z(1 ) p(1 p) < 보기 5.5.1> behavior of oral hygiee. 13 take oral exams per year out of % CI of p? p 13/ (0.59) / ,.46

26 5.6 두모집단비율의차이의신뢰구간 CI of differece of two proportios 100(1 - )% CI of p1-p? < 보기 5.6.1> p1 (1 p1 ) p(1 p) p1 p z(1 ) 1 Recovery times of a disease by two treatmets Assig 00 pts radomly to two trt groups. Trt A: 78pts recovered withi 3days, trt B 90 pts. 95% CI of p1-p? (.78)(.) (.90)(.10) (.78.90) 1.96 (.,.0)

27 5.7 평균을추정할때의표본수의결정 sample size calculatio: iferece of the mea (width of the CI)/, Samplig from a ifiite pop d ( reliability coef ) (SE) Samplig without replacemet from a small pop d z N N 1 d (reliability coef) (SE) z Nz d ( N 1) z z d

28 < 보기 5.7.1> measure daily protei itake from teeage girls. Width of CI=10 (+-5). Cofidece level= 0.95, pop sd=0, pop is very large; we ca igore fiite pop correctio factor z1.96, 0, d 5 z (1.96) (0) girls d (5)

29 5.8 비율을추정할때표본수의결정 sample size calculatio: iferece of the proportio < 보기 5.8.1> z pq d surveyig proportio of household with medical service. We kow p<0.35. d of 95% CI =0.05, =? Nz pq d ( N 1) z pq ( N 0.05) (1.96) (.35)(.65) households (.05)

30 숙제

31 5.9 정규분포모집단분산의신뢰구간 CI of the variace from ormal dist Poit estimator of variace Good estimator? ubiasedess Es ( )?

32 Pop=(6,8,10,1,14), = (exam at chap4) S ( xi ) N ( xi ) N w replacemet- w/o replacemet- si 0 0 Es ( ) 8 N 5 si 8 S N E( s ) 10 10

33 large N N N 1 E( s ) S ( 1) s CI of o the dis' of depeds = (x i-x) /. (chi-square distributio) i=1 ( 1) s 1 표 F

34 100(1 )% CI of? / (1 / ) ( 1) s ( 1) s 100(1 )% CI of ( 1) s ( 1) s (1 / ) / 100(1 )% CI of

35 < 보기 5.9.1> =15, measure Amylase, x 96, s35, 95% CI of? s 15, df 114 (14)(15) (14)(15)

36 5.10 두정규분포모집단의분산비에대한신뢰구간 CI for the ratio of two variaces 1 1 s s F 1, 1 1 표 G

37 1 100(1 )% CI of? s / (1 / ), s s F F s s s F F (1 / ) / < 보기 > ormal adults, =1(group1). Parkiso disease pts, =16명 (group). Observe respose time of a certai stimulus. Sample variace of grp1=1600,grp=15. 95% CI of? F.975 F s s s s,,

38 Statistical distributios : sum of idepedet ormal rv s 1,,, N(, ) Y Y Y radom sample from iid, Y N 1 1 ( 1) i i Y Y s 0,1 / Y N 1 / Y t s 1 (0,1) i i i Y Y N /, 1 1 /, 1 ( 1) 1 s P

39 : ratio of idepedet chi-squares (df= 1, ) F 1, 1 / / 1 F, 1 s s 1 / 1 / F 1, 1 1 PF 1 s1 / F 1 / s 1

40 Homework 종합문제