의학자료분석론

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의학자료분석론. 교재 : 강의록 Rosner B, Fundamentals of Biostatistics, 7 th ed. Brooks/Cole Cengage Learning, Canada, 2011. 강의 평가 : 출석 20% 숙제 30% 기말고사 50%. 의학연구에서의 통계학의 의의. 환자 진료 시 필요한 정보들 ( 예 : 진단방법 평가 , 정상치 판정 , 환자 모니터 등 ) 을 객관적으로 수집 , 분석하게 한다 . - PowerPoint PPT Presentation

Transcript of 의학자료분석론

: Rosner B, Fundamentals of Biostatistics, 7th ed. Brooks/Cole Cengage Learning, Canada, 2011.: 20% 30% 50%

1 (: , , ) , . . .2 . . . (by manual or by using the statistical packages). . .

3 (Descriptive Statistics) (Statistical Inference)4 (Descriptive Statistics) (Ways of organizing and summarizing a collection of numbers) (variable) (regularity) . , , (, )5 (Statistical Inference) ., . .6 (population) . (parameter) . (sample) (random sampling) (statistic) .

7(population) (sample)

2s2, ,8 sprbn

9 (nominal data) .) , O, A, B, AB , , C50.9(), C16.0() 2 (dichotomous or binary data) . 10 (ordinal data) . ) minor, moderate, severe . (categorical data) .11 (discrete data or interval data) . . (count) .) 8 1 , , .

12 (continuous data) . . ) ,

13 . . .

14 . . .

152. 16 . (Measures of location) . (mean) (median) (mode) . (range) (interquartile range) (variance) (standard deviation)

17 (Arithmetic Mean) . x1, x2, x3, . x10 . (mean) .

.

18 (median) 50 (50th percentile) .(central point)10 . 40, 120, 120, 125, 136, 150, 150, 150, 150, 167 . (n) , [(n+1)/2] . (n/2) [(n/2)+1] .19 10 . 5 6 .

. robust . 40 . 9 (9+1)/2=5 . 150/ .

20, (skewness)

.) 20 OC . )ModeMedianMean21 . : =Positively skewed: >Negatively skewed: 75th percentile + 1.5 * (75th 25th) Or 2) X < 25th percentile 1.5 * (75th 25th)(75th 25th) box . (adjacent value) .Box 3 (extreme outlying values) .

39Box plot .

403. Probability41 (probability) .1000 4 1000 5 . ? : . p-value .

42 (event)(event) (outcome) .50 5 . . (A, B, C )

43 A , A . - frequentist definition n m A . n m/n P(A) . ( ) .-empirical probabilities modeling . Goodness of fit .44 . 1992 4,065,014 , 2,081,287 . ,

Probability of an event A : P(A)0 P(A) 1

454. Discrete Probability Distribution46 (random variables) (probability distribution) () . ( ) . . .47 (Discrete Random Variables )n : (event, 1) / (non event, 0)xi : (X: )Pr(X=r) : sample N r

48Example 4.4 (p.82)100 4 . ,

. .() rPr(X=r) ( )00.0080.000=0/10010.0760.090=9/10020.2650.240=24/10030.4110.480=48/10040.2400.190=19/10049 0, 1, 2, 3, 4 5 (outcome) . outcome . outcome (probability mass function) . ( ) outcome .

(exhaustive events)

50 3 P(X=3) = 0.411 1 P(X=0) + P(X=1)= 0.008 + 0.076= 0.08451 .

52 outcome P(X=x) . 1. . . .(Goodness-of-fit test) .53() , . . (expected value) (population mean) . E(X) .

54 X ,

55, X Var(X) 2 . outcome .Var(X) xi .X .

56 (Binomial Distribution) Y . Y 2 . (Bernoulli random variable) .Y . Y=1, Y=0 . 30% . P(Y=1) = p = 0.3 P(Y=0) = 1 p = 0.7

57 . ?X .X = 0, 1, 2 3 .

P(Y1=1) = P(Y2=1) = 0.3 .XiY1Y2P(X=r)0001100121158P(X=0) = P(Y1=0 and Y2=0) : Y1 Y2 = (1-p)(1-p)= (0.70)2= 0.49P(X=1)= P( [Y1=1 and Y2=0] or [Y1=0 and Y2=1]= P(Y1=1 and Y2=0) + P(Y1=0 and Y2=1)= p(1-p) + (1-p)p= 2(0.30)(0.70)= 0.42P(X=2)=P(Y1=1 and Y2=1)=p2= (0.30)2= 0.09

59 X , X (binomial distribution) . Y n n outcomes .

p . X .X .n p (parameter) .

Y1Y2Y3..Yn010..160Sample of n independent trials : (n+1) discrete random variablesProbability of k event out of n trialsNumber of ways to select random variable kProbability of each outcome : each discrete random variableProbability of event at a trial ; a priori = p61 P(OXXOO) =(p)(1-p)(1-p)(p)(p)= p3(1-p)2P(OXOXO) =(p)(1-p)(p)(1-p)(p)= p3(1-p)2P(OOOXX) =(p)(p)(p)(1-p)(1-p)= p3(1-p)25 3 .

62N k ,

63N=5, P(X=2)=?5 2 ?P(boy) = 0.518 = pP(2 boys in specified 2 trials out of 5 trials) = (0.518)(0.518)(0.482)(0.482)(0.482)= (0.518)2(0.482)3# of ways selecting 2 trials out of 5 trials

= 0.3.

64 n p Rosner Book Table 1 . Table 1 n p k .65

66 2 . 5 2 .P=0.3, n=5P(X2) = P(X=0) + P(X=1) + P(X=2) = 0.1681 + 0.3602 + 0.3087 = 0.8370

67 10 7 .P(X7)= P(X=0) + P(X=1) + P(X=7)= 1 [ P(X=8) + P(X=9) + P(X=10)]= 1 [0.0014 + 0.0001 + 0.0000] = 0.9985

68 X

10 ,np = 10 x 0.3 = 3 .

,npq = (10)(0.3)(0.7) = 2.1 . .p=0.5 , p=0 or 1 .

69) 5% . 20 1 . ?, n=20, p=0.05 .20 , np = 20(0.05) = 1 .7020 1 ,

20 3 . 20 3 .

71 (unusual event?)20 3 ? P(X3) .

20 3 7.54% . 5% (unusual) . 20 3 .

72 n 20 p0.5 (normal approximation to binomial distribution) X ~ B(n, p) --> X ~ N(np, npq)p 0 1 .

73

74 Poisson Distribution . (rare event) . .

75) X 500,000 1 . p = 0.00024 .X n .

n .n p .

76 Rare event over time or over surface area

=t: expected number of events over the time period t : expected number of events per unit time or rate at which event occursk X .

77X 0 .t . 1 . 1subinterval of time 1 .78 vs. 1 .1 1 .1 , 1 , 1 p p 1 . 1 . .

n , .79

80 1. The probability that a single event occurs within a given small subinterval is proportional to the length of the subinterval.P(1 death)tP(0 death over t)=1-tP(more than 1 death over t)=0, 5 1 1 5.2. Stationarity : t t .3. Independence : .

81) 1 4.6 . 6 ? t=1 year, =4.6 = t 4.6=1, =4.6, 6mo=t=4.60.5=2.3

82Rosner book table 2 (p.822) . 0.5 20.0 0.5 .83 X

.

84 10 . (Normal approximation to poisson distribution)n p (n100, p0.01) . E(X) = np Var(X) = npq=(np)(1) = np E(X) = Var(X) --> Poisson approximation =np85) 500,000 1 X ,E(X) = = np= (500,000)(0.00024)= 120 120 . 50 ?

86 20 ?

0.05 . 20 (unusual) .

87 ? n 100 p 0.01 .() . .

885. --Normal DistributionThe most widely used distribution in statistical work.89 X . X (: DBP=117.3mmHg) =0 . . X . (: 90 X