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2 1 1 () () 1 random variable error

3 1 measurement data. cm (kg) (10 1 /L) / KPa count data ABABO 3 ordinal data population sample probability A PA 0PA n A m m/n A n frequency PA= m/n

4 3 sampling error [ ] A B C D D [ ] A B CD D 10 9 / L

5 1 ( ) A C A B. D. B. C D. 3 A B. C D. 4 A B. C D. 5 A 4

6 B C D population sample random sampling variation 4 (ordinal data) (probability) A PA PA A 0PA n A m m/n A n (freqency) PA= m/n random error (systematic error) 5

7 / L 10

8 7 X max X min R =

9 X X X = n X f fx X = f lg X G = lg 1 n f lg X G = lg 1 f M = X n+ 1 1 M = X n + X n ( ) ( + 1) R = X max X min P x 8

10 x P 5 ix P = L + ( n x% f ) x x L f x Q 3 = P 75 Q 1 = P 5 QR = Q3 Q1 ( ) X µ S = n 1 ( ) X ( X X X ) S = = n n 1 n 1 S CV = 100 % X S X ± 9

11 10

12 11

13 1

14 X ( X X) X 13

15 R = X max X min X f fx fx 14

16 ix n.0 M = Lx + fl fm 17 S = X = ( X X ) n 1 fx f fx (( fx) f ) f 1 X f 15

17 16 = f fx X + = L M x x f n f i L M = = = n X X ( ) X X = = n X X ( ) 1 = n X X S 100% S CV X = = = n X X ( ) 1 = n X X S 100% S CV X = = = = lg lg lg 1 1 f X f G

18 () 1 () () X 1 ( X µ ) (σ ) f ( X ) = e < X < + (3-1) σ π X X N( µ, σ ) µ σ µ σ µ σ µ x = µ µ σ σ σ σ σ σ µ = 0 σ = 1u Z u N1 17

19 µ = X u X N( µ, σ ) u σ u Φ( u) X, ) ( X 1 X 1 ( X µ ) ( σ ) D = e dx = Φ( u ) Φ( u ) 1 X1 σ π X1 µ X µ u1 = u = σ σ 3- X σ µ ± σ µ ± σ µ ± µ ±. 58σ ( X, X ) 1 % 90 X ± 1. 64S X 1. 8S X 1. 8S 95 X ± 1. 96S X 1. 64S X 1. 64S 99 X ±. 58S X. 33S X 33S + P ~ P P P P ~ P P P P 0.5 ~ P99. 5 P 1 P 99 X ± S X ± 3S t 18

20 + µ σ X = µ µ X µ σ X µ 1. 64σ µ σ µ σ µ +. 58σ µ ± 1. 64σ µ ± 1. 64σ µ σ X σ µ X A. X ±. 58S BX +.33S Clog 1 ( Y ±.58 S Y ) D log 1 ( Y +.33S ) Y X X Y X ( 10 1 / L) 1 ( 10 / L) 1 ( 10 / L) X X X X X 19

21 X µ P(4.00< X < 4.50) = P( < < ) 0.9 σ 0.9 = P ( 0.6 < u < 1.10 ) = 1 Φ( 1.10) Φ( 0.6) = = X 1 X 1.96σ = (0.9) = 3.61( 10 / L) 1 X σ = (0.9) = 4.75( 10 / L) ~ 4.75( 10 / L) 1P f L =0L X = 10 f X =7i X =1 1 P 0. 5 =10+ (4.98 0) = P f L =985 L = 18 X f X =8i X =1 1 P 95.5 =18+ ( ) = % 10.71~18.5 µ σ µ µ σ σ 0

22 u X ± S µ σ µ g /100g ( µ g /) l 1

23 X = S 38 ( µ g / g) 1 ( X µ ) (σ ) f ( X ) = e, < X < + σ π X ) N( µ, σ µ σ 3 N (0,1 ) µ = X u X N( µ, σ ) u σ ± X

24 X PX ( < 16.1) = P( < ) = Pu ( < 1.48) =Φ ( 1.48) = X PX ( >.9) = 1 - PX (.9) = 1 - P( ) = 1 - Pu ( 1.48) = Φ( ) = (14.6 P X 3.9) = PX ( 3.9) P( X 14.6) X X = P( ) P( ) = Pu ( 1.91) Pu (.13) =1 Φ ( 1.91) Φ (.13) = = x x 1 x 3 x P ( X > x ) = 0.05 Pu ( ) = x 19.5 P ( u 1.645) = = x = PX ( > x ) = 0.10 X 19.5 x 19.5 X 19.5 x 19.5 x 19.5 x 19.5 P P Pu ( > ) = ( < ) = ( < ) =Φ ( ) = 0.10 x 19.5 = 1.8 x.3 =.4 x 3 = 4.0 µ x n fx fx fx X = = = 1.15 µ n 00 3

25 µ fx ( fx) n (30) /00 S = = = n log 1 1 ( X SX ) = log ( ) = 39 µ µ X = ( µ g / l) S ) ( µ g / l 90% X S ( µ g / l) 95%X S ) ( µ g / l 99%X +. 33S ) ( µ g / l 4 90% P90 = 8+ (300 90% 67) =9.33 ( µ g / l) % P95 = 36+ (300 95% 85) =36.00 ( µ g / l) % P99 = 5+ (300 99% 97) =5.00 ( µ g / l) 9 X ± 1. 96S ± X ±. 58S ± ( µ g/ g) 4

26 4 1 sampling error standard errorse S S P S b X X σ X S X σ S σ σ = X n σ ( X = µ ) n S S X = n S= ( X X ) n 1 5

27 1 confidence intervalci CI 1-α CI CI 1-α α = % 95%CI α = % 99%CI 1-α () t t t t ν t (), 4-4- X S x X tα /, ν Sx X + tα /, ν Sx X uα / S x X + uα / S x X X 1 X tα /, ν S x 1 x X 1 X + tα /, ν S x 1 x () 1 t H µ = µ 0 0 H 1 µ µ 0 X µ t = 0 ν = n S X H µ = 0 H 0 d 1 µ 0 d t d µ d = = n 1 Sd ν 4- H0 µ 1 = µ H 1 µ 1 µ 6

28 X X t = ν = n + n S X 1 X 1 S ( ) ( ) c S = 1 1 n1 1 S1 + n 1 S X 1 X + S c = 4-4 n1 n n1 + n S X 1 X n>50 σ 0 X µ u = 0 n 4-5 S / n X µ 0 u = σ 0 σ / n 0 1 x 4-6 u = X 1 X 4-7 S x + S H 0 H0 H 1 α H 1 H1 H 0 β 4 β power of test α 1 A. B. C. D B [ ] S x = S / n n S x 7

29 X ±.58S x A99% B 99% C99% D 99% D 1 α CI 1 α BD A D 1 P=0.05 H P>0.05 H P=0.05 H 0 α P>0.05 α H 1 I II I II α 0 X S x 0 s ν = 14 8

30 1 0 g/l g/l H : µ = µ S x 1 x H : µ µ α = X X t = ν = n + n 1.7 t= = ν = P> mm/h t t= d / S = 3./0.611=5.37 d t= d / S = 5.0/0.948=5.303 d v=9p<0.001α =0.05 H 0 H 1 9

31 d t= d 1 S d 1 d =-1.60v=180.>P>0.1α =0.05 H /mm % 95% 1, X +1.96S= = /mm 3 X -1.96S= = /mm /mm 3 n=90>50 X +1.96S = / 90 = /mm 3 X X -1.96S X = / 90 = /mm /mm 3 6 AS BSE CS X DSD 7 A B C D 8 A B C t D 9 A B C D 10 A B t t C P 0.05 H 0 Du 6. Aα =0.05 Bα =0.01 Cα =0.10 Dα =

32 A B CAB DAB 8 At Bt Cu DF 9 30 X 1 S 1 X S AX 1 = X S 1 = S B t C F D 95% 10 A Bσ X Cσ X D 1 3 σ X P 7I II 8 9 α mmhg mmhg

33 mg% mg% mg% 4-9 3

34 4-9 mg% B 1 statistical inference sampling error σ X standard error of meansem confidence intervalci 1- α 1-α P H 0 I II I type I error H 0 I α II type II error H 0 II β 1- β power of test α H 0 H 1 level of a test α 1P P σ X 33

35 3 1P H 0 P 0.05 H 0 α H 0 α α α P H 0 1 t t=.484v=9p<0.05 α =0.05 H 0 t t=.157v=19p<0.05 α =0.05 H 0 α 95% 99% 95%99% %99% 0 34

36 1 1 3 LSD-t Dunnett-t SNK-q () 1 analysis of varianceanova sum of squares of deviations from meanss SS F 1 variation among groups, MS MS / SS SS = k i= 1 n ( x x) i i = ν ν =k-1 k (variation within groups) MS ) k ni MS = SS / ν SS = i= 1 j= 1 ( ( xij xi ), ν = N k 3 (total variation) 35

37 (MS ) MS = SS / ν SS = k n i ( i= 1 j= 1 N x ij x), k n i i ν =N-1 1 (homoscedasticity) 1 1 (completely random design) () 5-1F F α H 0 : µ 1 = µ = LL = µ k SS ν MS SS ν = k 1 ν SS ν ν ν ν ν MS MS 5-F Fα H 0: µ = µ = LL = µ 1 k SS ν MS ν SS ν MS MS 36

38 ν ν ν ν 37 ν ν SS SS ν ν MS MS t X X A B = d AB S d AB 1 1 S MS ( + ) = n A nb t x x i 0 = x i x 0 S xi x0 1 1 S MS ( + ) n i n 0 ( XA XB) MS 1 1 q = S = ( n + ) d A nb S d X = lg X X = lg( X + 1) Possion X = X X = X

39 X =1/ X 4 <30% >70% X = sin 1 X a 0 b 0 a 1 b 0 a 0 b 1 a 1 b 1 A B A B AB A B A AB 1 ASS <SS BMS <MS CMS =MS +MS D DSS =SS +SS [ ] SS =SS +SS C AB D H 1 H 1 38

40 t = F t = F A Bt C D ν Aν -ν Bν -ν Cν -ν +ν Dν -ν -ν 3 5% t s X µ t s A σ C 0.05, ν D. 0.05, ν x x 4 ν 1,ν α A B C D 39

41 5 A B C D 6 A B C D 7 A B C D 8 A B C D 9 SS A B3 C4 D33 10 t A B C DAB α α mg/l X ij

42 n i X i X ij s i X ij n i X i X ij s i Dunnett-t 5-5 % 41

43 n X i SS 16 SS = 8SS = 110 S = /mm ml/kg % MS 4

44 1C 1t t α P t α I H 0 α I H 1 α P H 1 P α H 0 H 0 C ij SS SS SS = ( X ) / n = /3 = = X ij C = = = [( X ) / n ] C = (167.9 = SS SS = ij i ) / = = SS ν MS F 43

45 F F =. 95 FF 0.05,3,8 0.05,3, 8 P<0.05=0.05 H 0 H 1 SNK-q H 0 A = B H 1 A B = X i q q P 1, <0.01 1, <0.01 1, >0.05, <0.01, <0.01 3, >0.05 P>0.05=0.05 H 0 4 P<0.01=0.05 H 0 H 1 H 0 H 1 =0.05 C ij SS = ( X ) / n = 88 / 30 = = X ij C = = 159. = [( ij) / i] = ( )/ = 41.6 SS X n C 44

46 SS = SS SS = = SS ν MS F F F F 0.05,, 7 = 0.05,,7 F P <0.05=0.05 H 0 H 1 Dunnet -t H 0 H 1 =0.05 Dunnett-t t = X X 1 0 MS (1/ n + 1/ n ) 1 = (9.-11.)/ 4.36(1/10 + 1/10) = /0.93 =.14 ν =7Dunnett-t P<0.05=0.05 H 0 H 1 t = ( ) / 4.36(1/10+ 1/10) =.99 ν =7Dunnett-t P<0.05=0.05 H 0 H 1 3 H 0 H 1 = SS ν MS F F F 3.88, P>0.05=0.05 H 0 = 0.05,,

47 4 = ( n i 1) si SS = (8 1) (5 1) 0.68 = ν = N - k = 35-5 = 30 + (5 1) 0.66 MS = SS / ν = /30= (8 1) (9 1) H H = t = ( + ) 8 5 =.16 ν = 30 =4 Dunnett -t.5t=.16<.5, P0.05 =0.05 H H 0 3 H 1 3 = t = = ( + ) 8 5 t>.5 P<0.05=0.05 H H 0 6 H 1 6 = t = = ( + ) 8 8 t>.5 P<0.05=0.05 H H 0 H 1 = t = = ( + ) 8 9 t>.5 P<0.05=0.05 H 0 5 N=36 k=3 n=1 46

48 SS SS SS SS = = 44 = v = N 1 = 36 1 = v = k v = n N v 35 1 = 3 1 = 1 = 1 1 = 11 = k n + 1 = = SS ν MS F P > > H 0 H 1 =0.05 H 0 H 1 = SS ν MS F F F 0.05,,10 = 4.10, F 0.05,4,10 = 3.48, P<0.05=0.05 H 0 7A, B SS ν MS F 47

49 * 19 F F 0.01,1,16 = 8.68, P<0.01=

50 () rate % proportion A 100% = 100% = B A =

51 ratio B A B = () 50 = % standardization method standardized rate adjusted rate (dynamic series) = n a n a0 = 1

52 51 1 4% A4% B4% C4% D4% B [ ] 100% =

53 %5% A. B. C. D. 1 A. B. C. D. 13 A. B. C. D /10 45/10 38/10 A. B. C. D. 15 a 0 a 1 a a n a + a a A 0 1 n B n+ 1 a0 a1 n + 1 a a n n C n n D 1 a a 0 A B 0 a n 5

54 C D 17 a 0 a 1 a a n a + a a A 0 1 n B n+ 1 a0 a1 n + 1 a n C a n n D 1 n a 0 a 18 C B D % 1/10 0~ ~ ~ a n 3. ~ ~1981 1/10 53

55 relative number rate = 100% % 3 proportion = 100% 4ratio AB A B A = B 5 (standardization method) 6 (dynamic series)

56 %1000 AB = / 100% % =/ 1/10 =/ 100% = 100% = = = n a n a

57 0~ ~ ~ ~ ~ 1/40~ 19.05%60~ 9.36%0~.90% 60~ / ~ % % 1/ N i P i N i P i P i N i P i

58 N i 64.4 N ip i N ip i N ' i Pi 771 p = 100 % = 100 % = 67.75% N 1138 N ' i Pi 648 p = 100% = 100 % = 56.94% N

59 Poisson 1 Poisson Poisson 1 Poisson Poisson X 0,1,,n P( X = k) = ( π ) n k n k ) π (1 (7-1) k X n 0,1,,n k µ µ P( X = k) = e >0 (7-) k! X Poisson XP 58

60 1 1 X = X σ = X X σ = n π ( 1 π) X XP X = X σ = X X X = µ Poisson X 1 X X k n i pi=1,,,k X=X 1X X k n pn=n 1+n ++n k X 1X X k i=1,,,k Poisson X=X 1 X X k = k Poisson Poisson n nπ ( 1 π) Poisson n nπ = λ Poisson Poisson P 1 n n p p-u p+u p ( 1 p) n 59

61 X Poisson S u p p 1 = S p p 1 60 X 1 + X X 1 + X 1 1 p = (1 )( ) n + n n + n n n p X X50 Poisson 1- X u X X + u X α α X Poisson k µ µ P( X = k) = e k=0,1,, k! X X Poisson u u X u 0 = u 0 u 3 Poisson u u

62 u X 1 = X X 1 + X u u 1 1 X 1 / n1 X / n = X n X + n () 1 A p=x/n Bn, Bn X Bn, C Bn, D Bn, B [ ] An Cn D 61 Bn Dn n nπ ( 1 π) P

63 10000 ± ± ± ± Y = X 1000 S Y S = X 1000 = P X 1 P = 1 (1 0.) + ( ) 0. (1 0.) = [ ] [ ] 80 = = X X 1X Bn 1p 1Bn p X 1X X= X 1 X AX 1=X B. n 1=n Cp 1 =p D. n 1 p 1 =n p 6

64 . Bnp An=50 B. p=0.5 Cnp=1 D. p=1 3. Poisson P µ N µ µ A µ µ B. =1 C µ =0 D. µ = % 3 A0.15 B C0.5 D A5195 C95105 B D Poisson Poisson 3 S p 1. 10% X 0,1,,n P( X π n k n k = k) = ( ) π (1 ) k X n Binomial Distribution 63

65 . 0,1,,n k µ µ P( X = k ) = e >0 k! X Poisson XP 3. Bernoulli A A Bernoulli Bernoulli Test n nπ ( 1 π) Poisson Poisson P 3. P n P S p 1. H 0 10% H 1 10% = % 3 3 P(X3)=1-[P(X=0)+P(X=1)+P(X=)]=1-[ *0.1* *0.1 *0.9 8 ]=0.070 P(X3)0.05=0.05 H 0. H 0 H 1 =0.05 S X + X 1 ( n 1 p 1 p = + n1 + n 1 1 ) n S p1 p = ( + ) =

66 u = p p 1 S p 1 p u = = P =0.05 H 0 H µ =100*4=400 µ Poisson H 0 400H 1 400=0.05 u = X u u u = = 30.58P = P( X = 0) P( X = 1) P( X = 10 ) 100 µ = 100 = Poisson µ P ( X ) = e 0! 0 1 µ µ µ µ µ + e 1! + e! = =

67 χ 1. χ. χ 1 χ 3 χ 3. χ χ 1χ () χ χ Chi-square test 1 3 () χ 1χ χ H 0 H 0 π 1 = π χ χ H 0 H 0 H 1 π 1 π ( ) A = T χ A Actual Frequency,T T Theoretical Frequency χ χ () 1 σ p π ( 1 π) = π (8-1) n p( 1 p) S p = p (8-) n n p 1-p p 66

68 p uα / S p, p + uα / S p (8-3) () χ χ H 0 H 1 H 0 H1 1 ( ad bc) n χ = ( a + b)( c + d)( a + c)( b + d ) n 40 1 T<5 ( ad bc n / ) n χ = ( a + b)( c + d)( a + c)( b + d ) ( b c 1) χ = b + c R C H0 ( H0 ) H1 ( H 0 ) (R-1)(C-1) A χ = n( 1) n R n C H 0 H1 ( A T ) T 1 n<40 χ χ 1 1<T<5n>40 T 1n 40 Fisher R C 1 <5 1/5 R C R C R C Kappa 1 χ A B C D 67

69 A [ ] χ χ χ χ χ + 0 χ + χ 1 χ 0 χ χ χ χ u χ 1 χ χ u = χ 1 χ R C χ R C χ χ H 0 H 1 α = 0.05 ( b c 1) = b+ c χ =.11< χ 0.05, ( ) χ = = ν = 1 P>0.05, 68

70 % H 0 H 1 α = 0.05 A χ = n( 1) = =4 n r n c ν =-13-1= α = 0.05 H 0 % χ H 0 H 1 α = 0.05 A χ = n( 1) =400 n r n c = ν =4-1-1=3 α = 0.05 H0 H 1 69

71 1. p A B χ Cn p 1-p p D t. A u B t C u = χ D t = χ 3. p ( 1 p) A p( 1 p) B n p n 1 4. χ A ν = n B 3 4 ν = 11 C ν =4 Dχ > χ ν > η 0.05, ν 0.05, η 70 p ( 1 p) n n % 50% 35% A B C D 6 A B χ C D 7 A u χ B u χ C u χ D χ u 1 χ Fisher

72 9McNemar 10Yates χ > χ 0.05(3) I II III % %

73 A B O AB Eskdale Annandale C χ χ α χ χ α χ 7 3 goodness of fit 4 A B n 8-1

74 8-1 A B a b - c d 5. R C 6. σ P σ p π ( 1 π) = π n π P π n p( 1 p) S p = n Fisher 1 n<40 R.A.Fisher1934 H 0 9. McNemar McNemar s test for correlated proportions ( b c 1) χ = v=1 b + c 10Yates Yates F χ χ 1 P χ correction of continuity Yates Yates correction H 1 H = χ. 37 P>0.05,,χ H 0: H 1 : α = 0.05 χ =61.59χ < χ 0.01, 6 P<0.05, 73

75 α = H 0 H 1, 3 χ H 0 : H 1 : α = 0.05 χ =3.00α = H 0 4 u χ χ χ =4.774α = H 0 H 1, 5R C χ χ =8.539v=P<0.05α = H 0 H 1 6R C χ χ = v=P<0.05α = H 0 H 1 7χ χ =4.00v=4P>0.05α = H χ χ =5.710 v=3p>0.05 α = H 0 74

76 1 non-parametric statistics distribution-free statistics assumption free statistics ( )

77 ( ) 7 (Wilcoxon ) H 0 M d H 1 M d α = 0.05 T + T-T + T- T + T- T n T P T P P n u u T n( n + 1)/ u = n( n + 1)(n + 1) / 4 u = T n( n + 1) / 4 n( n + 1)(n + 1) ( t 3 j 48 t j ) (Wilcoxon ) 1. (1)H 0 H 1 α = 0.05 () (3) n 1 T T (4) P T P T 76

78 P T P n 1 n -n 1 u u T n ( N + 1) / u = n n ( N + 1) /1 1 u c = u C 3 3 C = 1 ( t t ) ( N N ) j j (Kruskal -Wallis ) 1H 0 H 1 α = H 1 Ri H = ( ) 3( N + 1) N( N + 1) n H c i H c = H / C 3 3 C = 1 ( t t ) ( N N) j j 5 P H P T P T P (Nemenyi ) 1H 0 H 1 α = 0.05 D= R A -R B 3 P 1 D P 77

79 χ RA RB = C[ N ( N + 1)/1][1 n A + 1 n B ] 3 3 C = 1 ( t t ) ( N N ) j 1. j χ α,( k 1) (1) () R i (3) :R=b(k+1)/ b k (4) (R i -R); (5)M=(R ī R) (6)M M. Friedman (1) () R i χ k 1 = R bk( k + 1) j= 1 j 3b( k + 1) (3)χ α, ( k 1) P (1) R i () R R A B u = R A R B bk( k + 1)/ 6 u P c /c 78

80 A. B. C. D. D [ ] A.t B.u C. D.? C [ ] t H 0 A B C D B [ ] H H c C 79

81 [ ] Kruskal-wallis 3 3 Hc H c = H / C C = 1 ( t t ) ( N N) j C H CH P j j At B C D A B C D A B C D A B C D X AH 0 BH 0 80

82 CH 0 DH 0 AH 0 BH 0 CH 0 D At B C D A B C D A B C D n n i i i A B C D t u At u Bu t Ct u Dt u A B C D A B C D P 81

83 A < < B < C < D P n An Bn Cn Dn n 1 n An 1 n Bn 1 n Cn 1 n Dn 1 n A B C D A B C D X 1 X AX X BXX CX X DX X t A B C D A B C D H 0 A B Cµ d? DM d n 1 n n n 1 An 1 n 1 Bn n Cn 1n n 1n Dn 1 n n 1 n A B C 8

84 D At B Ct D t n 1n A B C D u c A B C D A B C D u c H 0 A B C D? 83

85 1 mol/l 84

86 mol/l () () 85

87 ? T - T + P n 1 T 1 n T P T - T + P t tp t n 1 T 1 n T P n 1 T 1 n T u c P 86

88 () 1. (linear regression) simple regression Y ˆ= a + bx ab ab a Y intercept regression coefficient X Y X Y b >0 a>0 b>0 Y X <0 a<0 b<0 Y X =0 a=0 b=0 X Y X ( X X )( Y Y ) lxy b = = a = Y bx ( X X ) l XX 87

89 . b 1 t () coefficient of product -moment correlat ion r. ( X X )( Y Y) r = ( X X ) ( Y Y ) = 88 l l XX XY l r 1r1 1 YY 0 r r r 1 3. r 1r t 1. 1 Y X XY 3b X Y b r 4b= l xy / l xx r = l xy / l l xx yy 5b+ 1r1 6b r.

90 1 b r, r b t t b =t r l xy 3 r = = SS SS l l xx yy r 1 rank correlation 1 A r B t C F D D [ ] t r b r b t t b=t r Ab 1 =b Ct r 1=t r D Bt b1 =t b D r b b 1 =b r b t r1 = t b 1 t r =t b t b1 =t b t r 1 =t r r 3 r >r 0.05( n-) X Y A B. C D. D r r r >r 0.05( n- ) P<0.05 H 0 H 1 0 X Y 4 H 0 A=0 B. 0 89

91 C>0 A D. <0 r r =0 r Ar>0b<0 Cr<0b>0 B B. r>0b>0 D. r b b r, r>0b>0 19 ( ) AXX BY Y /n C(YY ) DXX YY 0Y=14+4X 1~7 kg ( ) A C 1 r=1 ( ) Ab=1 CS Y. X =0 B D Ba=1 DS Y. X = S Y ( ) A B C D 3 X Y ( ) A µ S = ( X X ) ( n ) $ xy, B. Sr = ( Y Y) ( n 1) ( ) C. $ S = ( Y Y) ( n ) yx, D. S µ b = Sxy X X 4 ( ) An Bn1 90

92 Cn Dn1 5 Y ( ) ASS =SS CSS =SS 6 SS ( ) Al YY l XY b Bl YY bl XX YY XY XX C. l l l D(1 r ) l YY 7 ( ) BSS =SS D Al XY lxx lyy Bb YX lxx lyy Cb YX b XY D 8 r=0 Y ˆ= a + bx ( ) Aa B. a X Ca D. a Y 1 SS 1 =SS r 1 r 3 r 1 3 n=100x Y r=0.1 X Y 4r r s =0 95% cm cm

93 900g 95% 900g Y 95% g g a 10-4 a / ag/100m /L D 1 linear regression simple regression regression coefficient (slope) b b X Y b 3 residual sum of squaresss ( Y Yˆ ) X Y Y X 9

94 Y Yˆ 4 ( ) regr ession sum of squares SS ( Yˆ Y ) X Y Y X 5 linear correlation simple correlation 6 zerro correlation 9 coefficient of product -moment correlation r 30 coefficient of determination r SS l XY l XY l XX r = = = SS r l l l SS XX YY YY r 1 31 rectification 14. rank correlation 1 SS 1=SS SS SS r 1 r 3 r 1r1 r Y X a X ( ) ( ˆ) ( ) s Y. X s Y. X = SS n = Y Y n X Y Y X : b=0 y s Y. X Y X Y b = s l s b Y. X X Yˆ s s XX 1 n+ ( X X) ( X X ˆ = ) Y Y. X Yˆ s Y. X X Y y y 93

95 s ˆ = sy. X 1 n + ( X X ) ( X X) Y Y 3n=100r=0.1 t =0.05 H 0 (=0) H 1 (0) r =0.1 =0.01 1% 4 r r s 5 t 6 X YX Y Y ˆ= a + bx P X Y X Y 1~7 X Y Y X Y X X Y 7 X Y XY X Y X Y

96 1 X = 175 X = 9855 X = Y = 454 Y = 0690Y = 45.4 lxx = X ( X ) n = l = Y ( Y ) n = YY lxy = XY ( X )( Y ) n = r = l l XY l XX YY = 6 = = XY = = = = 6 H 0=0 H 1 0 r 0 t = = s r r ( 1 r ) ( n ) = = =0.05 ν = n = 10 = 8 t 0.00<P<0.005=0.05 H 0 H % r z 1 1+ r z = ln = ln = r z=tanh = z 95% 95

97 ( z u0.05 n 3 z + u0. 05 n 3) = ( ) = ( ) r=tanhz z 0 95% X=638X = X = 791 Y=173Y =06619Y = XY= l l l XX YY XY = = = l b = l XY XX Y X ( ( XY ( X ) Y ) X ) ( 1130 = = n = n = Y) n = = 1130 a = Y bx = = = = Y= X Y= X Y= X 10-1 H 0 0 H SS = l SS SS = l YY XY = SS = l XX SS = = = =

98 10-6 SS MS F F=16.147F P< H 0 H 1 t H 0 0 H SS SS SS s Y. X = l = l YY XY = SS = = l XX b 0 t = = s s b SS SS = = = = n = = Y. X b l XX 0.61 = = =6t 0.01>P> H 0 H 1 F = = = t Y ˆ = a + bx = X YYˆ X Y Yˆ YYˆ X=800Y=185 95% 95% 97

99 95% bt 0.05(n) S b bt 0.05(n ) S b = = X 1=690 Yˆ= XY 1=13.76X =934Y = % a = Y bx a 1 =78.85a = % 10- Yˆ= X Yˆ= X 900g 95% s Y = s 1 n + ( X X ) ( X X ) Y. X = ( ) = X=900 µ Ŷ 95% Yˆt 0.05(6) sy ˆYˆt 0.05(6) sy ˆ = = g g95% ~08.48g 900g Y 95% s Y = s 1+ 1 n + ( X X ) ( X X ) Y. X = ( ) = X=900 Yˆ= X= Y 95% Yˆt 0.05(6) S Y Yˆt 0.05(6) S Y = = g 95% 148.4~6.73g X 10-8 a 1/10 Y a d = d 98

100 d =8 H 0 s 0 a H 1 s0 a 0.05 r s =16d [n ( n 1)]=168[8(8 1)]= r s 0.10>P> H 0 a X /L Y d = d d =40.5 H 0 s 0 H 1 s Y r s r s T X =0 T Y=t 3 t 1=[(6 3 6)+( 3 )+( 3 )]1=

101 3 [( n n) 6] ( T + T ) X Y 3 3 [( n n) 6] T [( n n) 6] X 3 [( 1 1 ) 6] ( ) 3 3 [( 1 1 ) 6] 0 [( 1 1 ) 6] d r = s T = = Y 18.5 r s 0.10>P> H 0 100

102 P 6logistic SPSS SAS () Y Yˆ = b + bx + bx + + bx Yˆ X 1 X X k k b 0 b 1 b, bk partial regression coefficient b j X X j j Y () Yˆ X1 X X k Y b b 1, b k Yˆ Y e i = ( Y Y)ˆ b b 1, bk b1 b, bk normal equation k k 101

103 b l 1 b1l b1l 11 1 k1 + b l 1 + b l + b l k + Λ + b l + Λ k 1k + b l k k + Λ + b l l l ( X X )( X X ) XX l k kk = l 1 y = l = l y ky ( X )( X ) i = = = ij ji i i j j i j iy = ( X i X i )( Y Y ) = ( X Y b0 b = Y b X b X Λ i X i )( n b X k k n Y ) j X1 X X k Y H 0 β 1 = β = β 3 = L = β k = 0, H1 β j 0 0 F MS lk F = = MS l( n k 1) n k l = b l + b l + Λ + b l 1 1y l = l l y ( Y Y ) l = l () logistic = yy X, X, Λ, X 1 k Y Y Y =1Y k ky Y =0P Q P + Q=1 Logistic Q P e 1+ e β0+ β1x1+ βx + Λ + βk X k = β + β X + β X + Λ + β X 0 = β + β X + β X + Λ + β 1+ e β 0 β j ( j= 1 L k) X j P x β x β ( ) P Q 01 n i X, X, Λ, X Y i1 i ik i k k X k k 10

104 Y =1 Y i i =0 P Q i i P i + Q i =1P i Q i l P i β + β + β + L + β e = 1 + L e 1 = + L X X X 0 1 i1 i k ik β + β X + β X + + β X 0 1 i1 i k ik Q i X X X 0 1 i1 i k ik 1 e β + β + β + + β i odds P Q l l i i P Q OR odds ratio Pi Q i ln = 1( X i1 X l1 ) + ( X i X l ) + + k ( X ik X lk ) Pl Q β β Λ β l ( ) j = 1Λ k X X ( ) ij X lj i X ij X lj β j X j X j logistic logistic maximum likelihood estimate β b ( j= 1 L k) j j Y X, X, 1 Λ, X k 1i Y i = 0 i i l = P Q Yi 1 Yi i i i n L n n Y = = i 1 Yi L li Pi Qi i= 1 i= 1 i 1 n L Newton Raphson e β j 103

105 ( j = 1Λ k ) b j logistic logistic likelihood ratiotest (score test)wald (Wald test) Λ = ln ' ' ( L L) = (ln L ln L) ln L m ( m < k) L ' m 1 X H χ Λ χ α(1) α X j j X j 0 1 Y D E F D D [ ] X Y j SS SS R R = SS SS 104 R p R Logistic logistic logistic SPSS SAS

106 X 1 /kg X /cm X 3 /cm Y /L SPSS EXAP11sav 4 0 Statistic Regression Linear DependentY Independent(s)X 1 X X 3 MethodEnter Variables Entered/Removed Model Variables Entered Variables 1 X3, X, Removed Method. Enter 105

107 X1 a All requested variables entered. b Dependent Variable: Y Model Summary Model R R Square Adjusted RSquare Std. Error of the Estimate a Predictors: (Constant), X 3, X, X 1 ANOVA Model Sum of df Mean F Sig. Squares Square 1 Regression Residual E- 0 Total a Predic tors: (Constant), X, X, X 3 1 b Dependent Variable: Y Coefficients Unstandardized Standardized Model Coefficients Coefficients t Sig. B Std. Error Beta 1 (Constant) X 3 X X E E E a Dependent Variable: Y SAS DATA EXAP11INPUT x1 x x3 PROC REG CARDS MODEL y=x1 x x RUN 106

108 Analysis of Variance Sum of Mean Source DF Squares Square F Value Pr > F Model Error Corrected Total Parameter Estimates Parameter Standard Variable DF Estimate Error t Value Pr > t Intercept X X X [ ] SPSS SAS 1 H 0 β 1 = β = β 3 = 0 F=13.413P= H 0 b 1b b 3 Yµ = X X X R =0.715R a = SE t P b b b b A χ B F C U 107 D Ridit 33 A B C D

109 34 k B C 1/k D E 35 F A B C D logistic β i Y kg X 1 cm X 1 X1 =1611 X1 = X =106 X = 976 Y = 341 Y = 9883 X 1 X = X 1Y = X Y = Y ˆ= b0 + b1 X1 + b X b =.114 b = b = v SS MS F 1B A 3B 4A 1 Y X multiple linear regression multiple regression Yˆ = b 0 + bx bx + + bx b k k 1 b, bk partial regression coefficient b j X j X j Y 3 R coefficient of multiple correlation, R 108

110 4 coefficient of determination R SS SS R = SS / SS R y x 5logistic β0+ β1x1 + βx + Λ + βk X k P = β0+ β1x1 + βx + Λ + βk X 1+ e Q e = β + β X + β X + Λ + β 1+ e i oddsp Q i i k k X k odds 6logistic OR odds ratio β j X j j X j 1 Y ˆ= b0 + b1 X1 + b X b1 l11 + bl1 = l1y b l + b l = l 1 1 y b = b = b = Y ˆ= X X v SS MS F e β 109

111 110 1 (statistical graph) 1 1

112 111

113

114 113

115 114 [ ] 1-

116 115

117

118 117

119 118 (1)

120 119

121 10

122 (%) 11

123 1. 1

124 13 1-π () (experimental study) experiment clinical trial

125 14 error bias 1 random error systematic error selection bias measurement bias contamination intervention compliance noncompliance lost of follow-up disagreement blind method

126 confounding bias randomized block design π ( ) σ u α + u β N = 13-1 δ N u β δ N = uα 13- σ u β 1- β 15 β

127 N ( u u ) σ + α β d N = 13-3 δ u β N = δ uα 13-4 σ d 3 ( u α + u β ) σ N = 13-5 δ N uβ = δ uα 13-6 σ 4 N = ( u + u ) α β ( π π ) 1 4π 1 π c c ) 13-7 u β = N π π π c 1 ( 1 π ) c u α 13-8 π 1 π π c N 5 χ uα π c + uβ π + π+ N = π + π + b c = π = + a + b a + c N π + π + uα π u β = π π π π c c π c π c π = π u = α σ N δ 7 π( 1 π ) = u α N 13-1 δ 5. Kappa PA Pe Kappa = P e Kappa Kappa

128 experimental effect [ ] 1. [ ]. [ ] 1. 10% 5% 0.05, α = 90% α u 0.05 = 1.64β u π π = δ π c 1 ( u + u ) 4 π (1 π ) α β c c ( ) N = = = 946 ( π π ) g 467g 0g α = β = β α =

129 ( ) µ α + µ β σ 1 ( ) N = = =111 δ 0 11 ( ) µ α + µ β σ ( ) N = = =154 δ δ N u β = uα = 1.64 =. 00 σ 467 µ β =.00 β =0.0power=1-0.0= ~64 5.5% 5.0% 1α = % α =0.05 (1) α =0.05,u α β uβ π1 π π c N = ( u + u ) 4π ( 1 π ) α β ( π π ) 1 c c = ( ) ( ) ( ) = 7588 α = % α =0.05,u α β uβ N = ( u + u ) 4π ( 1 π ) α β ( π π ) 1 c c = ( ) ( ) ( ) α =0.05, u α β u β N = ( u + u ) 4π ( 1 π ) α β ( π π ) 1 c c = ( ) ( ) ( ) α =0.05, u α = 3694 N π 1 π u β = uα = 1.96 = π c ( 1 π c ) 0.05( ) β = = = % 10% 0% α β α u 0.05 = 1.96π 0 δ 18

130 uα 1.96 N = π0(1 π0)( ) = 0.1 (1 0.1) = 864 δ α =0.05,β α u 0.05 = 1.96π 0 δ uα 1.96 N = π0( 1 π0) = ( ) = 4838 δ A. B. C. D. 50 A. B. C. D. A. B. C. D. A. B. C. D. A. B. C. A. C. D. B. D. 19

131 A. 130 B. C. D. A. B. C. D.. E A 8 E A A 1 A. B. C. D. A. t B. C. t D. 3 A. H0 A H A 1 B. H0 A H1 A C. H A 0 H1 A D. H 0 A H1 A 4 A. P0.05 A B. P0.05 A C.P0.05 A D.P0.05 A

132 1. A 1.8ml/minB.4ml/min 1.0ml/min =0.05= III α β kpa 10 5kPa α % ABC ABC placebo control randomization confounding factor systematic error bias (experimental study) 131

133 lost of follow-up randomized control trial D. D 3. C 4. A 5. D 6. D 7. D 8. BC, A 1B A 3D 4D =.4-1.8=0.6ml/min =1 =0.05 =0.1 u = u 0.01 = 1.8 t ( ) σ ( ) u α + u β N = = =10 δ α u 0.05 = 1.64 β u 0.05 = 1.64 δ 10 0 Cσ 0 C t ( u + u ) α β ( ) 0. N σ + = = = 43 δ α u 0.05 = 1.64 β u 0.01 =.33 δ kpaσ 5kPa σ t ( u + u ) α β ( ) 5 N σ + = = = 98 δ α u 0.05 = 1.96 β u 0.05 = 1.64δ 6σ 6 t ( uα + uβ) σ ( ) 6 N = = = 51 δ

134 π = c 3 α u 0.05 = 1.96 β u 0.10 = 1.8π 1 = π = ( u ( ) 4 (1 ) α + uβ)4 πc(1 π ) + c N = = = 85 ( π 3 1 π) ( )

135 13-1 A B C 140 ABC ABC χ χ A B 6 48 C , 134

136 135 () 1 (1) () (3) (4) (5) () (survey) ( ) () 1

137 136

138 t / σ n = ( α ) δ δ σ n ν t 0.05 / u t 0.05 / 0.05 / n u α / p(1 p) n = δ δ p p p p p p p () 1 (probability sampling) Sampling frame) 1 simple random sampling) 137

139 n S = ( 1 ) 14-3 N n S X n p(1 p) S p = (1 ) N n n / N (sampling fraction) ( 1 n / N) (finite population correction) (systematic sampling) 3 (stratified sampling) strata) (proportional allocation) ni Ni n = n = N 14-5 i i n N N (optimum allocation) Niσ i ni = n 14-6 N σ i i Ni πi (1 πi ) ni = n 14-7 N π (1 π ) i i i W = N N X i i / = i X W i X 14-8 n = i S X (1 ) Wi S N i p = W i pi n = i S p 1 ) Wi N X i ( S pi i (cluster sampling) K k 138

140 X K X m X i = i Nk 14-1 k K k 1 S X = ( 1 )( ) ( Ti T ) N K k( k 1) i= 1 T i i T T i T = Ti / k K p = a i Nk k K k 1 S p = ( 1 )( ) ( ai a ) N K k ( k 1) i a a i= 1 (non-probability sampling) 1 (quota sampling) 30% 30% (snowballing) (judgement) () (case control study) (cases) (controls) (cohort study) (exposure) (outcome) 139

141 14-1 (odds ratioor) (cumulative incidenceci) OR RR OR RR (relative riskrr) (attributable risk proportionar%) (incidence densityid) (attributable riskar) (attributable risk proportionar%) ORRR () ( ) ( ) (standardization) 1 (direct standardization) ( ) (indirect standardization) ( standardized incidence ratiosir standardized mortality ratiosmr)sir(smr) 140

142 n 1 r 1 p 1 N 1 R 1 P 1 n r p N R P 3 n 3 r 3 p 3 N 3 R 3 P 3 i n i r i p i N i R i P i k n k r k p k N k R k P k n r p N R P N p i i p = N r p = P n P i i r SMR = n P i i (survey) ( ) [ ] 1 A B C D 141

143 A [ ] A B C D D [ ] 3 A C [ ] 4 A B C D A [ ] 5 14

144 [ ] n k K k [ ] (kpa) %.(kPa) σ =.δ =0.5 u 05 =1.96 t σ n = ( α / ) n ν t 0.05 / δ u0.05/ t 0.05/ n u σ n = ( α / ) n δ n ν ν t t 0.05/,73 143

145 t / σ n = ( α ) δ n ν ν t t n = ( t α / σ ) δ 0.05/,75 t / σ n = ( α ) δ u σ n = ( α / ) δ 1 A C B D n A δ σ β B α σ µ C δ σ α D δ σ µ 3 (5000 ) 0.5mg/L 0.1mg/L A97 B95 C96 D94 4 A B C D 5S = S / n x ACD 6 144

146 7 8 9 A B C D 10 A B 1 13 C 1 D A B C 400 D % 0%α mg/L 0.1mg/L ( ) ( ) 0~ ~ ~ ~ ~

147 OR ( a ) 5(b ) 135( m 1 ) 140( c ) 5( d ) 365( m 0 ) 50( n 1) 50( n ) 0 500(n ) 1 (sampling survey) simple random sampling) 3 (systematic sampling) 4 (stratified sampling) strata) 5 (cluster sampling) K k 6 (probability sampling) 7 (non-probability sampling) 146

148 134B 56A 7B 1011A 8C 9D 1 1 p =0.1δ =0.p =0.0.1=0.0 u 05 =1.96 α / u p(1 p) n = = / δ σ =0.5δ =0.1 u =1.96 σ t α n = ( / ) n ν t 0.05/ δ u t 0.05 / 0.05 / n u σ n = ( α / ) n δ n ν ν t t n = ( t α / σ ) δ 0.05 /,

149 ~ ~ ~ ~ ~ N i pi p = = = / 10 N ~ ~ ~ ~ ~ r p = P 140 n P = 13.1 = / 10 i i OR = = OR Mantel-Haenszel χ H OR 1 0 H OR 1 1 ( ad bc) ( n 1) MH χ = =73.17ν =1 n n m m

150 73.17> χ 0.05, 1=7.88P <0.05 H

151 150 1 () (population) = 65 =

152 14 = (pyramid) = (general fertility rategfr) = 15~49 = = Σ( ) natural increase rate, NIR = 151

153 15 standardized mortality ratesmr IMR = 50% 100% = 8 =

154 153 = () incidence rate n n n n

155 154 Kaplan-Meier infant mortality rate, IMR = [ ] 1 A B C D B [ ]

156 155 A B C D D = ( ) A B C D C A B CD A,

157 = 100 =

158 157

159 158

160 159 (pyramid) = = standardized mortality ratesmr

161 160 n n n n Kaplan-Meier

162 161

163 () current life table cohort life table 1 5 () 1 (exact age) age specific probability of dying X n d X n d X q X = n q X = 16-1 lx lx q X X n q X X n 16

164 d X n d X XX+n 3 number of survival person-years l X lx d X n d X qx n q X d = l q d = l q 16- X X X n X X n X lx+1 = lx d X l X+ n = lx n dx number of survival person-years total number of survival person-years X n LX n L X l X X X+n L 0 = l1 + a0 d a 0 0 lw L w = 16-5 mw L w l w m w 5 life expectancy X T e = x 16-6 x l x p i n X p i n X r i x p i = ( p ) n (16-7) n X n X 1 lx n d X n q X e X X 163

165 1 3 4 life expectancy X X X G 65.5 H 65.5 I 65.5 D D [ ] X X+1 1q x X+1 X+ 1q x+1 X X+ ( ) A 1q x 1q x+1 B1-1q x1q x+1 C1-1 q x 1 q x +1 D q x 1 q x+1 ( ) 164

166 A 0 B 0 C 0 D m85 ( + ) e ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ a 0 = life table current life table 3 complete life table 4 abridged life table 5 5 cohort life table 6 age specific probability of dying X n 165

167 7 number of survivors X 1D D 1 m = D / p X n q = n m ) /( + n m ) n X ( n X n X n qx = mx /[ 1+ (1 ax ) mx ] a X XX+1 X m 85 ( + ) e () 1 n DX 1 n m X = 16- P n q n X 46 q q = = X 16-3 d = l q X l d n X l = = = l 1 = l0 d0 = = = l q = = l = l d = = 1 1 d n X (4) L n X a = L = l + a d = L = l 41 = ( + ) = = m80( + ) n X

168 (5) T L X = n X n L X T L = = ( + ) 75 = L75 + T80 = T T X (6) e X = lx T e 0 = = = l T e 1 = = = l X ~ 1 P n X D n X m n X 4 q n X 5 l X 6 d n X 7 L n X 8 T X 9 0~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ e X

169 1 Cox 1 survial time 11-1 t 360,990,1400, complete data censored data = = /9/80 11/04/ /13/8 06/08/ /0/83 1/31/ /04/83 04/10/ mortality probability q q = 168

170 1 survival probability p p = 1 q = Kaplan-Meier 1 1 survival rate t ) S t k t S ( t k ) = P( T tk ) = k 11-1 T () p 1, p, Λ, p k S ( t k ) = P( T tk ) = p 1 p Λ pk 11- p, p, Λ, p 1 k 0 t k 1 product -limit method Kaplan-Meier 1958 Kaplan-Meier life-table method log-rank test m ( Ak Tk ) χ = υ = m T k= 1 k A k T k k H 0 χ m 1 χ m χ H 0 Cox Cox ( k 169

171 Cox h t ) = h ( t ) exp( b X + b X + Λ + b X ) p 11-4 ) h(t ( p h 0 ( t ) X, b Cox D R b Cox i 1 ( ) A B C B D B C D 3Cox ( ) A 170 B C D Cox Cox DCox h ( ) AB C 0 t 4 log-rank ( ) A B C D D

172 Kaplan-Meier k () t t n t d q = d n p =1 q k S ( t k ) SE( S ( t k ) ) ( 3 4 )= ( 3 4 )( 3)= ( 3 4 )( 3)( 0 1)= k =1 1 t ) 3 t n t 3 4 t d t 4 5 t 5 6 t

173 1Cox ( ) A B C D A B C D 3 ( ) A C B Db,c survival analysis survival time 3 complete data 17

174 4 censored data 5 mortality rate 6 mortality probability 7 survival rate tk t k 8 survival probability 1 h 0 () t k k () p = 1 q S ( t k ) SE( S ( t k ) ) t d c n 0 n c = - c n 0 q = d n (7) (8) (9) (10) 1 0~ ~ ~ ~ ~ ~ ~ ~ ~ ~

175 11 10~ ~ k =1 1 n c = n 0 - c 6 3 q = d n 7 4 p = 1 q t a 11- log-rank 1 H 0 H 1 α = A =4A 0 1 =10T 0 =8.6694T 1 = m ( A ) ( ) ( ) k Tk χ = = + = T k = 1 k 4 P 1 χ P <0.05α = H0 H 1 174

176 175 evaluation synthetical evaluation (pre-event evaluation) (medial evaluation) (after-event evaluation)

177 176 (system s analysis method) (specialist-scored method) Satty Topsis

178 1 % 3 % % 7 8 % 9 % 10 % 11 % y = X M M y = X X M 177

179 I = m n y ij i= 1 j= 1 I = = I = = A C B D 178

180 A B C D Saaty A B C D A C B DTopsis 5 m 1 I = y n 1 Am Cy Bn D 18-5 Topsis 1995~ ~1997 % % % % % % % % A.B 3.D 4.C 5.C % % % % % % % %

181 X ij Z = 18-4 ij n ( X ij ) Z11 = = i= Z = a a, Λ a ( i1 max, imax, immax) Z = ( a, a, Λ, a ) i1 min i min im min + Z = ( 0.60,0.604,0.601,0.578,0.580,0.580,0.580,0.580,0.583,0.581 ) Z = ( 0.538,0.535,0.560,0.577,0.576,0.574,0.575,0.576,0.57,0.57) D D 18 8 D D + i i 1997 D + D D + = = = = m ( a ij aij ) i= max 18-7 m ( aij a ij ) i= 1 min 18-8 ( ) + ( ) + Λ + ( ) ( ) + ( ) + Λ + ( ) D = = C C i Ci = = i Di C i = 18-9 D + D + i i

182 + D D C i