populatio sample II, B II? [1] I. [2] 1 [3] David J. Had [4] 2 [5] 3 2

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1 (2015 ) 1 NHK !? New York Times For Today s Graduate, Just Oe Word: Statistics Google Hal Varia I keep sayig that the sexy job i the ext 10 years will be statisticias SF H.G. Wells Statistical thikig will oe day be as ecessary for efficiet citizeship as the ability to read ad write. Advaced Theory of Statistics M. 2 a) b) c) d) 17 e) f ) g) h) i) j ) χ 2 1

2 populatio sample II, B II? [1] I. [2] 1 [3] David J. Had [4] 2 [5] 3 2

3 2 variable descriptive statistics A, B, C A, B, C * =1: 0: class frequecy 2 4 relative frequecy 5 6 cumulative frequecy cumulative relative frequecy 3

4 histogram 1 Frequecy Test score averages mea 4

5 arithmetic mea *1 x 1, x 2,, x x = x 1 + x x = 1 i=1 x i x = = (1, 1, 1, 1, 2, 3, 4, 5, 16, 20) media x (1), x (2),, x () = 2m + 1 m + 1 x (m+1) = 2m m m + 1 (x (m) + x (m+1) )/2 = 10 5 x (5) = 2 6 x (6) = mode rage *1 x G x H x G = x 1 x 2 x, ( ) 1 = x H x 1 x 5

6 Iter Quartile Rage (IQR) 50% 100p%(0 p 1) 100p quartile Q 1 25% 2 Q 2 50% 3 Q 3 75% Q 3 Q 1 50 Q 1 = 47.25, Q 3 = 69.5 IQR= variace stadard deviatio x 1, x 2,, x x i x (i = 1, 2,, ) 2 S 2 S 2 = 1 { (x1 x) 2 + (x 2 x) (x x) 2} = 1 S 2 = 1 * 2 (x i x) 2 = 1 i=1 x 2 i x 2 i=1 (x i x) 2 2 S S = S 2 = 1 (x i x) i=1 i=1 2.4 box-ad-whisker plot Q 1 Q x i (i = 1, 2,, ) x S z i = x i x S *2 1/ 1/( 1) 6

7 z i (i = 1,, 2,, ) 0 1 z z z i = cotigecy table cross table i(= 1,, ) 1 (x i, y i ) x y x y scatter plot

8 3 50 correlatio coefficiet (x 1, y 1 ), (x 2, y 2 ),, (x, y ) x y covariace S xy = 1 (x i x)(y i ȳ) i=1 x y ( x, ȳ) (x i x)(y i ȳ) x i, y i 1 3 x i, y i 2 4 (x i, y i ) x S x y S y r r = S xy S x S y = 1 1 i=1 (x i x)(y i ȳ) i=1 (x i x) 2 1 i=1 (y i ȳ) x y z i = (x i x)/s x, w i = (y i ȳ)/s y r xy = 1 z i w i 1 r xy 1 1 i=1 8

9 x y 3 z (1, 1) (1, 2) (1, 3) (6, 5) (6, 6) Ω ω Ω = {(1, 1), (1, 2), (1, 3),, (6, 5), (6, 6)} evet A, B, 2 2 A A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} 1 φ {1} {1, 3, 5} 5 2 A, B A B A B A, B A B A B A B A B = φ. 1 A B 3 A B = {1, 2, 3, 5} A B = {1, 3} C A C = φ A C 9

10 3 A, B, C (A B) C = (A C) (B C) (A B) C = (A C) (B C). 2 A 2 B 2 C 2 5 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} B = {(1, 1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5, 1), (5, 3), (5, 5)} C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} A B = {(1, 1), (1, 3), (1, 5), (2, 2), (3, 1), (3, 3), (3, 5), (4, 4), (5, 1), (5, 3), (5, 5), (6, 6)} (A B) C = {(1, 1), (1, 3), (2, 2), (3, 1)} A C = {(1, 1), (2, 2)}, B C = {(1, 1), (1, 3), (3, 1)} (A C) (B C) = {(1, 1), (1, 3), (2, 2), (3, 1)} 1 2 A A A c φ c = Ω, Ω c = φ A A c = Ω, A A c = φ Ω A = A, φ A = A (A c ) c = A A, B (A B) c = A c B c, (A B) c = A c B c. 1 A B 3 A B = {1, 2, 3, 5} (A B) c = {4, 6} A c = {2, 4, 6}, B c = {4, 5, 6} A c B c = {4, 6} (a), (b), (c) (a) A 0 P (A) 1 (b) P (Ω) = 1 (c) A 1, A 2, A 3, P (A 1 A 2 A 3 ) = P (A 1 ) + P (A 2 ) + P (A 3 ) + 10

11 N A R A P (A) P (A) = R N A A A α P (A) = α 100 (a), (b), (c) A B 3 A B c, A B, A c B A B = (A B c ) (A B) (A c B) (c) P (A B) = P (A B c ) + P (A B) + P (A c B) A A B c, A B A = (A B c ) (A B) (c) P (A) = P (A B c ) + P (A B) P (B) = P (A c B) + P (A B) P (A B) = [P (A) P (A B)] + P (A B) + [P (B) P (A B)] = P (A) + P (B) P (A B) P (A B) = P (A) + P (B) P (A B) A B A B = φ P (A B) = P (A) + P (B) 11

12 A A c = φ, Ω = A A c (a), (c) 1 = P (Ω) = P (A) + P (A c ) A = Ω A c = φ P (φ) = A B 3 A = {1, 3, 5}, B = {1, 2, 3} P (A) = 3/6 = 1/2, P (B) = 3/6 = 1/2 A B = {1, 3} P (A B) = 2/6 = 1/3 P (A B) = P (A) + P (B) P (A B) = = 2 3 A B = {1, 2, 3, 5} P (A B) = 4/6 = 2/ A B 5 A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}, B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)} P (A) = 6/36 = 1/6, P (B) = 10/36 = 5/18 A B = {(1, 1), (2, 2)} P (A B) = 2/36 = 1/18 P (A B) = P (A) + P (B) P (A B) = = = 7 18 A B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)} P (A B) = 14/36 = 7/ A B A B P (A) = 1/5 3 1/3 B A P (A B) B B A 1, 3,

13 P (A B) B A P (A B) = P (A B) P (B) P (B) = 3/5, P (A B) = 1/5 P (A B) = 1/5 3/5 = 1 3 P (A B) = P (B)P (A B) A B A 2 B 2 A B 1 P (A) = 3/ P (B A) = 2/9 2 P (A B) = P (A)P (B A) = = 1 15 A B P (A B) = P (A)P (B) A B P (A) = P (A B) P (B) = P (B A) A B A 2 1 B 1 A B P (A B) = 1/36 P (A)P (B) = (1/6) (1/6) = 1/36 A B 3.2. (i), (ii) (i) B 1 B 2 = φ P (B 1 B 2 A) = P (B 1 A) + P (B 2 A) (ii) P (B A) + P (B c A) = (i) A 6 B 4 A B (ii) C 7 B C P (A B) P (B A) 3.5 A = A Ω = A (B B c ) = (A B) (A B c ) A B A B c (c) P (A) = P (A B) + P (A B c ) 13

14 P (A) = P (A B)P (B) + P (A B c )P (B c ) P (B A) = P (A B) P (A) = P (A B)P (B) P (A B)P (B) + P (A B c )P (B c ) A B 1 B c 2 P (B) = P (B c ) = 1/2 P (A B) = 3/4, P (A B c ) = 1/3 P (B A) = (3/4)(1/2) (3/4)(1/2) + (1/3)(1/2) = 9 13, P (1/3)(1/2) (Bc A) = (1/3)(1/2) + (3/4)(1/2) = 4 13 P (B c A) P (B A) + P (B c A) = 1 2. A B P (A B) = 0.95, P (A c B c ) = 0.90 P (B) = P (B A) P (B A) = (1 0.90) ( ) = (1) (2) 99% 99% 99% 99% 99.1% 99.9% ) sesitivity specificity (1) 3 1% (2) 2%

15 /6 radom variable X, Y, X P (X = 1) = 1/6, P (X = 2) = 1/6,, P (X = 6) = 1/6 X x X X. 2 X X 2, 3,, 12 P (X = 2) = 1/36, P (X = 3) = 2/36, P (X = 4) = 3/36, P (X = 5) = 4/36, P (X = 6) = 5/36, P (X = 7) = 6/36, P (X = 8) = 5/36, P (X = 9) = 4/36, P (X = 10) = 3/36, P (X = 11) = 2/36, P (X = 12) = 1/36 X 6 X x 1, x 2,, x K f(x 1 ), f(x 2 ),, f(x K ) X f(x k ) = P (X = x k ) (k = 1,, K) X f f(x k ) 0 K f(x k ) = 1 k=1 X 1 X a b P (a X b) = X f b a f(x)dx f(x) 0 15 f(x)dx = 1

16 X P (X = a) = 0 0 X x F (x) = P (X x) X cumulative distributio fuctio F (x) = x f(u)du F (x) = f(x) F (X) = u x f(u) 4.2 expectatio E(X) E(X) = xf(x), E(X) = x xf(x)dx x x X g(x) E(g(X)) = g(x)f(x), x E(g(X)) = a, b g(x)f(x)dx E(aX + b) = ae(x) + b a = 0 E(b) = b b = 0 E(aX) = ae(x) X x 1,, x K K K K K E(aX + b) = (ax i + b)f(x i ) = a x i f(x i ) + b f(x i ) = ae(x) + b i= X i=1 E(X) = = 21 6 = /2 X 2 E(X 2 ) = = X k 3C k (1/2) k (1/2) 3 k = 3 C k (1/2) 3 (k = 0, 1, 2, 3) ( ) 3 ( 1 1 E(X) = 0 3 C C ) C 2 ( 1 2 i=1 ) 3 ( ) C 3 = / X X 16

17 /2 X µ = E(X) variace V (X) = E[(X µ) 2 ] V (X) 0 V (X) X V (X) stadard deviatio σ 2 σ V (X) = x (x µ) 2 f(x), V (X) = (x µ) 2 f(x)dx V (X) = E(X 2 ) 2µE(X) + µ 2 = E(X 2 ) µ 2 = E(X 2 ) (E(X)) 2 a, b V (ax + b) = a 2 V (X) a = 0 V (b) = 0 X x 1,, x K K V (ax + b) = K {(ax i + b) (aµ + b)} 2 f(x i ) = a 2 i= E(X) = 7/2, E(X 2 ) = 91/ E(X 2 ) = C 0 ( 1 2 ) C 1 ( 1 2 V (X) = 91 6 K i=1 ( ) 2 7 = ) C 2 ( 1 2 (x i µ) 2 f(x i ) = a 2 V (X) ) 3 ( ) C 3 = = V (X) = 3 ( ) 2 3 = X 4.3. X { 1 (0 x 1) f(x) = 0 (x < 0, 1 < x) 17

18 X 1 Y X, Y 0, 1 X = x, Y = y f(x, y) = P (X = x, Y = y) f(0, 0) = , f(1, 0) =, f(0, 1) =, f(1, 1) = 2 X, Y X Y f(x, y) X, Y f(x, y) 0 f(x, y) = 1 x y f(0, 0), f(1, 0), f(0, 1), f(1, 1) 0 f(x, y) = f(0, 0) + f(1, 0) + f(0, 1) + f(1, 1) = 1 x=0,1 y=0,1 X, Y f(x, y) f(x, y) 0 f(x, y)dxdy = 1 X = 0 P (X = 0) = P (X = 0, Y = 0) + P (X = 0, Y = 1) = f(0, 0) + f(0, 1) X = 1 Y = 0 Y = 1 g(x) = P (X = x), h(y) = P (Y = y) g(x) = f(x, 0) + f(x, 1), h(y) = f(0, y) + f(1, y) X, Y X, Y g(x) = y f(x, y), h(y) = x f(x, y) X, Y g(x) = X, Y P (X = x Y = y) = f, g, h f(x, y)dy, h(y) = f(x, y)dx P (X = x, Y = y) P (X = x, Y = y), P (Y = y X = x) = P (Y = y) P (X = x) g(x y) = f(x, y) f(x, y), h(y x) = h(y) g(x) h(y) 0, g(x) 0 g(x y) y x h(y x) x y 18

19 x, y f(x, y) = g(x)h(y) X Y X, Y X, Y g(x y) = g(x)h(y) h(y) = g(x), h(y x) = g(x)h(y) g(x) = h(y) X Y Y X 2 X, Y X, Y X, Y X h(y) 1 1/36 1/36 1/36 1/36 1/36 1/36 1/6 2 1/36 1/36 1/36 1/36 1/36 1/36 1/6 Y 3 1/36 1/36 1/36 1/36 1/36 1/36 1/6 4 1/36 1/36 1/36 1/36 1/36 1/36 1/6 5 1/36 1/36 1/36 1/36 1/36 1/36 1/6 6 1/36 1/36 1/36 1/36 1/36 1/36 1/6 g(x) 1/6 1/6 1/6 1/6 1/6 1/ (a), (b), (c) (a) 2 (b) p (c) X x F x f(x) = P (X = x) = C x p x (1 p) x (x = 0, 1,, ) biomial distributio B(, p) C x p x (1 p) x = (p + (1 p)) = 1 x= X X B(10, 0.5) f(x) 7 f(x) = 10 C x 0.5 x x (x = 0, 1,, 10) 19

20 X /6 1 1 X B(10, 1/6) f(x) 8 f(x) = 10 C x ( 1 6 ) x ( ) 10 x 5 (x = 0, 1,, 10) 6 7 B(10, 0.5) 8 B(10, 1/6) X B(, p) E(X) = p, V (X) = p(1 p) 5.1. B(, p) X E(X) = p, V (X) p(1 p) 5.2 p p x f(x) = C x p x (1 p) x p = λ, p 0 f(x) = e λ λ x x! (x = 0, 1, 2, ) Poisso distributio P o(λ) e x e x = 1 + x + x2 2! + x3 3! + = e λ λ x = e λ x! x=0 X P o(λ) x=0 r=0 x r r! λ x x! = e λ e λ = 1 E(X) = λ, V (X) = λ P o(λ) λ B(, p) E(X) = p, V (X) = p(1 p) X λ = 2 P o(2) 20

21 5.3 ormal distributio { } 1 f(x) = exp (x µ)2 2πσ 2 2σ 2 ( < x < ) σ > 0 E(X) = µ, V (X) = σ 2 µ σ 2 N(µ, σ 2 ) N(µ, σ 2 ) x = µ x = µ x ± f(x) 0 µ x σ N(20, 2 2 ), N(20, 4 2 ), N(20, 6 2 ) X N(µ, σ 2 ) X Z = X µ N(0, 1) 0 1 σ stadard ormal distributio f(z) Φ(z) f(z) = 1 2π exp ) z ( z2, Φ(z) = 2 Φ( z) = 1 Φ(z) 1 2π exp ) ( x2 dx 2 Φ(z) Excel Φ(1.645) 0.95, Φ(1.96) Z = X µ N(0, 1) Z k k (k > 0) σ P ( k Z k) = P (Z k) P (Z < k) = Φ(k) Φ( k) k = 2 P ( 2 Z 2) = Φ(2) Φ( 2) ( ) = 0.95 Z 95% 2 2 X X 95% µ 2σ µ + 2σ 1. X N(9, 25) P (3 X 21) Z = X 9 5 N(0, 1) ( 3 9 P (3 X 21) = P X ) = P ( 1.2 Z 2.4) = Φ(2.4) Φ( 1.2) z = 2.4 Φ(2.4) = Φ( 1.2) = 1 Φ(1.2) Φ(1.2) = Φ( 1.2) = P (3 X 21) = =

22 2. 17 N(172, 4 2 ) cm 17 % X X N(172, 4 4 ) P (X 180) ( ) X 172 P (X 180) = P 2 = 1 Φ(2) = cm 2.5% 5.3. T T N(50, 10 2 ) 40 T 60, T 70, T 75, T 55, 50 T BMI (Body Mass Idex, kg/ m 2 ) 22.65(kg/m 2 ) 4.01(kg/m 2 ) 20 BMI N(22.65, ) 20 BMI 25(kg/m 2 ) % B(, p) X N(p, p(1 p)) B(, p) N(p, p(1 p)) 1 X B(, 1/6) /6, 5/36 10, 20, 30, 50, 100 B(, 1/6) 10 B(, 1/6) X B(, p) ( ) a p P (X a) Φ p(1 p) p > 5 (1 p) > 5 p = 1/2 10 p X X 100 X B(720, 1/6) Z = X ( ) X 120 P (X 100) = P 2 = P (Z 2) < > [1] 22

23 6 6.1 populatio sample X 1, X 2,, X X 1, X 2,, X X 1 X 1 X 2 *3 X 1, X 2,, X f(x) sample size (i) (ii) 2 (i) X 1 X 2,, X µ σ (ii) % *3 23

24 6.3 µ σ 2 µ = xf(x), σ 2 = (x µ) 2 f(x) x x µ = xf(x)dx, σ 2 = (x µ) 2 f(x)dx µ σ 2 X 1, X 2, X sample mea sample variace X 1, X 2,, X f(x) X = X 1 + X X X, Y E(X + Y ) = E(X) + E(Y ) E( X) = E(X 1) + E(X 2 ) + + E(X ) = 1 i=1 X i = µ + µ + + µ = µ = µ X, Y V (X +Y ) = V (X)+V (Y ) V ( X) = 1 2 V (X 1 + X X ) = σ2 + σ σ 2 X 0 N X µ = N µ s 2 = 1 1 {(X 1 X) 2 + (X 2 X) (X X) } = S 2 = 1 {(X 1 X) 2 + (X 2 X) (X X) } 2 = σ2 (X i X) 2 s 2 1 E(s 2 ) = σ 2 s 2 σ 2 S 2 E(S 2 ) = 1 σ2 s 2 S 2 = 10 1 i=1 6.4 X X/ X/ X/

25 X X µ, σ 2 X i (i = 1,, ) X X N(µ, σ 2 ) X = (X X )/ N(µ, σ 2 /) B(, p) i (i = 1,, ) 1 0 X i X i B(1, p) S S = X X E(X i ) = p, V (X i ) = p(1 p) S N(p, p(1 p)) µ σ 2 µ σ 2 θ ˆθ µ ˆµ = X X X 1 = x 1, X 2 = x 2,, X = x θ ˆθ X µ s 2 σ 2 25

26 p x X X B(, p) p P (X = x) = C x p x (1 p) x p L(p) = C x p x (1 p) x p p L(p) dl(p)/dp = 0 p ˆp = x/ L(p) log L(p) d log L(p) = 1 dl(p) dl(p) = 0 dp L(p) dp dp d log L(p) = 0 dp 1 N(µ, σ 2 ) µ 1. ˆθ θ ˆθ θ E(ˆθ) = θ X = (X X )/ µ s 2 σ 2 1/ S 2 2. θ 7.2 θ 1 α P (L θ U) 1 α L, U L, U 1 α [L, U] 100(1 α)% cofidece iterval α 0.01, 0.05, % 95% 90% % 95% % 95 θ 90% % 90 θ µ p 95% 26

27 7.2.1 µ N(µ, σ 2 ) X 1, X 2,, X µ i) X N(µ, σ 2 ) ax + b N(aµ + b, a 2 σ 2 ) ii) X, Y N(µ 1, σ 2 1), N(µ 2, σ 2 2) X +Y N(µ 1 +µ 2, σ 2 1 +σ 2 2) X Y N(µ 1 µ 2, σ σ 2 2) X 1, X 2,, X N(µ, σ 2 ) ii) X 1 + X X N(µ, σ 2 ) X = (X 1 + X X )/ i) N(µ, σ 2 /) X σ/ (a) X µ (b) 1/ X µ 2 σ/ 1/ σ/ 1/ X N(µ, σ 2 /) X Z Z = X µ σ/ Z N(0, 1) σ 2 X N(0, 1) N(0, 1) Φ(z) = 1 α z Z α Φ(z) = z Z Φ(z) = 0.95 z Z P ( Z α/2 X ) µ σ/ Z α/2 = 1 α µ ( ) σ P X Z α/2 µ X σ + Z α/2 = 1 α σ 2 µ 1 α 100(1 α)% [ ] σ σ X Z α/2, X + Zα/2 95% [ X 1.96 σ, X σ ] * σ 2 s 2 = 1 i=1 1 (X i X) 2 ( X µ)/s 1 t t σ

28 5 X = ( )/5 = 68 95% [ X 1.96 σ ] [ σ 240.7, X = , ] = [54.4, 81.6] = % [ X 1.96 σ ] [ σ 245.6, X = , ( ) t 49 t [53.9, 62.8] ] = [54.0, 62.7] BMI ( kg/ m 2 ) 22.65(kg/m 2 ) 4.01(kg/m 2 ) 20 BMI 95% 7.2. σ 2 = 9 µ 95% p B(1, p) X 1, X 2,, X p p ˆp = X = i X i/ i X i B(, p) p i X i N(p, p(1 p)) ( i X i p)/ p(1 p) P ( Z α/2 i X i p p(1 p) Z α/2 ) 1 α ˆp = i X i/ p ( p(1 p) P ˆp Z α/2 ) p(1 p) p ˆp + Z α/2 1 α p ˆp p P ( ˆp Z α/2 ˆp(1 ˆp) ) ˆp(1 ˆp) p ˆp + Z α/2 1 α p 100(1 α)% [ ] ˆp(1 ˆp) ˆp(1 ˆp) ˆp Z α/2, ˆp + Z α/ B(1, p) = p 95% A 100 A p 95% 95% 40% ±5% 28

29 X B(10, 0.5) 1 P (X = 1) = 10 C = P (X 1) = P (X = 0) + P (X = 1) = % 1.07% 0.5 hypothesis testig 8.2 1) H 0 H 1 α 2) p 3) p < α H 0 H 1 sigificat p α 10 p 0.5 H 0 : p = 0.5 ( p 0.5) H 1 : p < 0.5 H % α 5% 10% p 10 p P (X 1 H 0 ) =

30 p α p α 10 α = < α = > µ µ 0 α σ 2 H 0 : µ = µ 0 H 1 : µ µ 0 H 1 : µ > µ 0 H 1 : µ < µ 0 (X 1, X 2,, X ) X N(µ 0, σ 2 /) Z = X µ 0 σ/ N(0, 1) Z Z 0 0 α Z > Z α/2 Z Z α/2 Z z p P ( Z z H 0 ) α X * σ 2 s 2 µ 0 s/ 1 t t t σ µ = µ 50 5% H 0 : µ = 50 H 1 : µ X = ( )/5 = 68 z = / % Z α/2 = Z 0.05/2 = 1.96 z = > 1.96 p P ( Z 2.594) = P (Z 2.594, Z 2.594) = P (Z 2.594) + P (Z 2.594) = Φ( 2.594) + 1 Φ(2.594) = 2[1 Φ(2.594)] 2( ) = µ BMI (Body Mass Idex, kg/ m 2 ) 22.65(kg/m 2 ) 4.01(kg/m 2 ) 20 BMI 23kg/m 2 5% = 257 z = /

31 z = < 1.96 p P ( Z 1.399) = P (Z 1.399, Z 1.399) = P (Z 1.399) + P (Z 1.399) = Φ( 1.399) + 1 Φ(1.399) = 2[1 Φ(1.399)] 2( ) = % BMI 23kg/m 2 * 2 2 BMI 23kg/m 2 23kg/m = % p p p 0 α H 0 : p = p 0 H 1 : p p 0 (X 1, X 2,, X ) X i (i = 1,, ) 0 1 B(1, p 0 ) S = X X B(, p 0 ) N(p 0, p 0 (1 p 0 )) Z = S p 0 p0 (1 p 0 ) = ˆp p 0 p0 (1 p 0 )/ ( ˆp = S ) N(0, 1). 50% 100 5% H 0 : p = 0.5 H 1 : p p Z = /100 = 2.0 P ( Z 2.0) = 2[1 Φ(2.0)] 2( ) = /100 = = % 31

32 8.3 2 α 2 i) 1 (α ): ii) 2 (β ): 2 2 α 1 α 5% 1 5% p α 1 α β power 32

6.1 (P (P (P (P (P (P (, P (, P.

6.1 (P (P (P (P (P (P (, P (, P. (011 30 7 0 ( ( 3 ( 010 1 (P.3 1 1.1 (P.4.................. 1 1. (P.4............... 1 (P.15.1 (P.16................. (P.0............3 (P.18 3.4 (P.3............... 4 3 (P.9 4 3.1 (P.30........... 4 3.