2 1,, x = 1 a i f i = i i a i f i. media ( ): x 1, x 2,..., x,. mode ( ): x 1, x 2,..., x,., ( ). 2., : box plot ( ): x variace ( ): σ 2 = 1 (x k x) 2
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1 1 1 Lambert Adolphe Jacques Quetelet ( ) (1 ) x 1, x 2,..., x ( ) x a 1 a i a m f f 1 f i f m 1.1 ( ( )) x f x 1, x 2,..., x 1. mea or average ( ):,. x = 1 x k k=1
2 2 1,, x = 1 a i f i = i i a i f i. media ( ): x 1, x 2,..., x,. mode ( ): x 1, x 2,..., x,., ( ). 2., : box plot ( ): x variace ( ): σ 2 = 1 (x k x) 2 = 1 k=1 x 2 k x 2 stadard deviatio ( ): σ = σ 2 = 1 (x k x) 2 x, σx, 2 σ x., σ 2 = 1 (a i x) 2 f i = (a i x) 2 f i = a 2 f i i x2 i i i 1.2 Iferetial Statistics k=1 k=1 Statistics Experimets Measuremets Data Statistical Iferece Good decisio Useful iformatio Probability Theory
3 1.3. What is a Radom Variable? What is a Radom Variable? (radom variable)., X, Y, Z, T,.... Discrete radom variables ( ) (1) 3. (2). Cotiuous radom variables ( ) (1) 1,. (2). X. X. X,. 1.4 Distributios of Discrete Radom Variables 1.2 3, X. X {0, 1, 2, 3}., P (X = 0) = 1 8, P (X = 1) = 3 8, P (X = 2) = 3 8, P (X = 3) = 1 8,.. x P (X = x) 1 8 X,, X ( ). X {a 1,..., a i,... } x a 1 a i P (X = x) p 1 p i
4 4 1, P (X = a i ) (, ),. p i = P (X = a i ), p i 0, p i = 1. (p i = 0 a i, p i = 0.) i E[X] = m X = i a i p i = i a i P (X = a i ), V[X] = σ 2 X = i (a i m X ) 2 p i = i a 2 i p i m 2 X., ( ). V[X] = E[(X m X ) 2 ] = E[X 2 ] E[X] ( ) E[X] = 3 2, V[X] = Distributios of Cotiuous Radom Variables 1.3 R 1, X. X [0, R]. a X = a P (X = a) = 0,. R x F (x) = P (X x). x < 0 F (x) = 0, x > R F (x) = 1., 0 x R. X x 1 O x, O x. 1,., F (x) = P (X x) = πx2 πr 2 = x2 R 2. 2 x, 0 x R, f(x) = R2 0,. X.
5 1.5. Distributios of Cotiuous Radom Variables 5 X, ( ) f(x) = f X (x). F X (x) = P (X x) f X (x),, F X (x) = P (X x) = : x f(x) 0, P (a X b) = f X (t)dt d dx F X(x) = f X (x). + b a f(x)dx = 1. f(x)dx, a < b, f (x) E[X] = m X = + + a b x xf(x)dx, + V[X] = σx 2 = (x m X ) 2 f(x)dx = x 2 f(x)dx m 2 X., V[X] = E[(X m X ) 2 ] = E[X 2 ] E[X] ( ) E[X] = 2 3 R, V[X] = 1 18 R2. HW 1 2, X,,. HW 2 2, L, S., L = S. L, S,,. HW 3 L X. (1) X. (2) (1), 2. (3) X.
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7 7 第 2 章 基本的な離散分布 Sime o-deis Poisso ( ) 2.1 Biomial Distributio (二項分布) 表が出る確率が p であるコインを 回投げたとき, 表の出る回数 X の分布 ( ) k P (X = k) = p (1 p) k, k = 0, 1, 2,..., k を二項分布といい, B(, p) で表す. 特に, B(1, p) を成功確率 p のベルヌーイ分布という. 例 題 2.1 B(4, 1/2) と B(4, 1/4) を図示せよ. k k P (X = k) P (X = k) 定 理 2.2 二項分布 B(, p) の平均値と分散は m = p, σ 2 = p(1 p) 確率母関数 {0, 1, 2,... } に値をとる確率変数に対して pk = P (X = k) (k = 0, 1, 2,... ) とお く. このとき, pk xk f (x) = k=0 を X のまたは確率分布 {p0, p1,... } の母関数という. 補 題 2.3 確率母関数について次が成り立つ. (1) f (0) = p0, f (1) = 1. (2) E[X] = f (1). (3) V[X] = f (1) + f (1) {f (1)}2.
8 Geometric Distributio ( ) p, X P (X = k) = p(1 p) k, k = 0, 1, 2,.... p. ( ) 2.4 p m = 1 p p, σ 2 = 1 p p Poisso Distributio ( ) X λ > 0, P (X = k) = λk k! e λ, k = 0, 1, 2, λ m = λ, σ 2 = λ. 2.6 ( ) B(, p) p = λ ( ),, p 0, λ. 2.7 ( ) 1 3,., 1, (1) 1. [0.05] (2) 5. [0.18] HW ,. 5, ( ). [199.8 ] HW 5, 20., 1 3. HW ? 1 365,, 5 5 X B(50, 1/365)., P (X = k) (k = 0, 1, 2, 3, 4). [ , , , , ; : , , , , ]
9 2.3. Poisso Distributio ( ) 9 HW 7 X λ. (1) P (X = 0) P (X = 1) λ. (2) X, P (X = k) k.
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![11 第 3 章 基本的な連続分布 Joha Carl Friedrich Gauss (1777 1855) 3.1 Uiform Distributio (一様分布) 区間 [a, b] からどの点も同等な確からしさで 1 点を選ぶときのモデルとして現れる. 1, a x b f (x) = b a 0, その他 定 理 3.1 [a, b] 上の一様分布の平均値と分散は, m= 3.](/docs-images/93/111303398/images/11-0.jpg "2 a+b, 2 σ2 = (b a)2 12 Expoetial Distributio (指数分布) ランダム到着の待ち時間をモデル化するときに現れる. λ > 0 を定数として { λe λx, x 0 f (x) = 0, x<0 定 理 3.2 パラメータ λ の指数分布の平均値と分散は, m= 3.")
11 11 第 3 章 基本的な連続分布 Joha Carl Friedrich Gauss ( ) 3.1 Uiform Distributio (一様分布) 区間 [a, b] からどの点も同等な確からしさで 1 点を選ぶときのモデルとして現れる. 1, a x b f (x) = b a 0, その他 定 理 3.1 [a, b] 上の一様分布の平均値と分散は, m= 3.2 a+b, 2 σ2 = (b a)2 12 Expoetial Distributio (指数分布) ランダム到着の待ち時間をモデル化するときに現れる. λ > 0 を定数として { λe λx, x 0 f (x) = 0, x<0 定 理 3.2 パラメータ λ の指数分布の平均値と分散は, m= 3.3 1, λ σ2 = 1 λ2 Normal Distributio (正規分布) N (m, σ 2 ): 平均 m, 分散 σ 2 の正規分布 (またはガウス分布) { } (x m)2 1 exp f (x) = 2σ 2 2πσ 2 定 理 3.3 (de Moivre Laplace の定理) 二項分布は, 同じ平均と分散をもつ正規分布で近似 できる. B(, p) N (p, p(1 p)), 0 < p < 1,.
12 B(100, 0.4) N(40, ) Stadard Normal Distributio ( ) N(0, 1) ( ) X N(m, σ 2 ), ax + b N(am + b, a 2 σ 2 ),, Z = X m N(0, 1) σ 3.6 Z N(0, 1). (1). P (Z 1.15), P (Z 1.23) [0.8749, ] (2) a. P (Z a) = 0.33, P (Z < a) = 0.75 [0.44, 0.67] (3) X N(2, 5 2 ), P (X 0) , 225 ( ( ) ). HW 8 500, 250. HW 9 ( ) m, σ, ( ) = x m σ,., 20 80,. HW 10, 4% ,. [0.0901]
13 3.4. Stadard Normal Distributio ( ) N(0, 1) 13 I(z) = 1 2π z 0 e x2 /2 dx z
14 ,. 30cm 40cm, 5cm. [1/2] , 10.,. 2.,. [9/25] 3 3 3, 4, 5 1 P, P ( 5 ) 1. [95/144] 4 ( ) X, F (x) = F X (x) = P (X x)., x. 3, X. X,. 5 L Y. Y,,,. [F Y (x) = 0 (x < 0); = 2x/L (0 x L/2); = 1 (x > L/2). f Y (x) = 2/L (0 x L/2); = 0 (otherwise). E[Y ] = L/4. V[Y ] = L 2 /48.] 6 R 1, X. X. [E[X] = R/3. V[X] = R 2 /18.] 7 O R 1, O X. X,,,. 8 (, ) λ. [.] 9 N 4. 1 N N, 4, X. 4 k N, P (X = k), E[X]. [4(N + 1)/5] 10 (1) X N(20, 4 2 ), P (X > 17.8). [0.7088] (2) X N(50, 10 2 ), P (X > a) = a. [28.3] 11 60, 1 12., 600, 1 120,. [.] 12, 5%., 68, 8.,. [ ]
15 15 4 I Jacob Beroulli ( ) 4.1 Samplig ( ) ( ),, ( ). 1, x 1.,, x., X, x X. Radom Samplig with Replacemet ( ) 1,.,, 1 X 1, 2 X 2,..., X., X 1, X 2,..., X ( ). Estimate of Populatio Parameters ( ),..,,.. X 1, X 2,..., X,.,,.
16 16 4 I 4.2 Poit Estimatio, f(x 1, X 2,..., X ) (poit estimatio)., X = 1 k=1 X k ( ) ( ) E[ X] = m. 4.2 ( ) X, ( ) P X = m = 1. lim 4.3 (Strog law of large umbers ( )) X 1, X 2,..., m., ( P lim 1 k=1 ) X k = m = ( ), 1, 0., x 1, x 2,... t = 1 k=1 x k. t,
17 4.3. Biomial Populatio Biomial Populatio E, 2, E p..,, E 1, E 0. m = p. X 1, X 2,..., X. k, X k = { 1, k E, 0, k E,..,,, ˆp = 1 X k., ˆp. 4.5 (Audiece Ratig Survey ( )).,? ( ) 5 1 ( ) ( ) 10 (%) 16/04/27( ) 8: /05/01( ) 20: /05/01( ) 21: /04/27( ) 22: /04/27( ) 21: /04/30( ) 21: /04/25( ) 21: /04/30( ) 20: /04/28( ) 21: /04/25( ) 21: /04/29( ) 12: k=1 32,., 27, PM.,, PM 600, 200. ( ) : ( )
18 18 4 I 4.4 Iterval Estimatio of Biomial Parameter ˆp, (!),., ˆp p., ˆp,. (iterval estimatio). ˆp. (1) X k B(, p). k=1 (2), B(, p) N(p, p(1 p)) ( ). p 5, (1 p) 5. (3), ˆp = 1 X k N k=1 ( p, ) p(1 p) ˆp p p(1 p)/ N(0, 1) (4) 2 ( ), p ˆp : ˆp p ˆp(1 ˆp)/ N(0, 1). α = α/2 α, Z N(0, 1) ( ) P ( z Z z) = 1 α z N(0, 1) α. z α α N(0,1) 1 α α/2 α/2 -z 0 z
19 4.4. Iterval Estimatio of Biomial Parameter 19 p 1 α [ ] ˆp(1 ˆp) ˆp(1 ˆp) ˆp(1 ˆp) ˆp z, ˆp + z ˆp ± z. 90% (α = 0.1, z = 1.64) 95% (α = 0.05, z = 1.96) 99% (α = 0.01, z = 2.58). α 1 0 (1 α) 0% 100% 0 ( ) ( ) ( ), x 1..., x (, x k = 0 = 1). ˆp,.,.., 1 α, α.,. 4.6 ( ) %. 95%, 0.141( ) ± ± , 95% 0.01,? [38416] HW , 51% (NHK ). 90%. [0.51 ± 0.027] HW 12, 90% 0.02,? [6724] HW 13,.
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![21 5 II William Sealy Gosset (1876 1937) 5.1 5.1 ( ) X, Y, E[XY ] = E[X]E[Y ], V[X + Y ] = V[X] + V[Y ] 5.2 ( ) N(m, σ 2 ) X 1, X 2,..., X X = 1 X k, X N ) (m, σ2 k=1 X m σ/ N(0, 1) m, σ 2,,.](/docs-images/93/111303398/images/21-0.jpg "( ) ( ) P X = m = 1. lim 5.3 ( ) X 1, X 2,..., m = 0, σ 2 = 1., ( ) lim P 1 X k x = 1 x e t2 /2 dt. 2π,, k=1 1 X k N(0, 1). k=1 5.2 ( ) X 1, X 2,.")
21 21 5 II William Sealy Gosset ( ) ( ) X, Y, E[XY ] = E[X]E[Y ], V[X + Y ] = V[X] + V[Y ] 5.2 ( ) N(m, σ 2 ) X 1, X 2,..., X X = 1 X k, X N ) (m, σ2 k=1 X m σ/ N(0, 1) m, σ 2,,. ( ) ( ) P X = m = 1. lim 5.3 ( ) X 1, X 2,..., m = 0, σ 2 = 1., ( ) lim P 1 X k x = 1 x e t2 /2 dt. 2π,, k=1 1 X k N(0, 1). k=1 5.2 ( ) X 1, X 2,..., X : m ( ), σ 2 ( ) m 1 α, X ± z σ z N(0, 1) α (= α/2 ) (5.1)
22 22 5 II p 1 α, ˆp ± z ˆp(1 ˆp) (5.2)., (5.1)., p p(1 p). (5.2), (5.1), σ 2 ˆp σ 2 = ˆp(1 ˆp). 5.4, 200, 2.2 g., 1.5 g., g?. [95% 2.2 ± 0.208] HW g., 8g. 1. [95% 156 ± 2.48] HW 15 HW14, 95% 1g? [984] 5.3 ( ) X 1, X 2,..., X : m ( ), σ 2 ( ) U 2 = 1 1 (X i X) 2, S 2 = 1 i=1 (X i X) 2 i=1,. (,, ) : E[S 2 ] σ U 2 : E(U 2 ) = σ 2.,, S 2 U N(m, σ 2 ) X 1,..., X, T = X m U/ t 1 ( 1) t-,.
23 5.3. ( ) 23 t- 1 B ( 2, 1 2) ( ) t2 2 = Γ( +1 2 ) Γ( 2 )Γ( 1 2 ) ( ) t2 2 (5.3) (1) Γ. Γ(x) = 0 t x 1 e t dt, x > 0. (2) B. B(x, y) = 1 (3) N(0, 1),. 0 t x 1 (1 t) y 1 dt = Γ(x)Γ(y), x > 0, y > 0. Γ(x + y) (4) = t- N(0, 1). (5), 30 N(0, 1). m 1 α, X ± t U t t 1 α 5.7 8,. 90% [ x = , u 2 = = , t 7 = ± 0.375] HW 16,. 95%. [33 ± 4.17] HW 17 (5.3), = t- N(0, 1). [Γ(1/2) = π.]
24 24 5 II t ( α P ( T t (α)) = α) \α α 0 t ( α)
25 25 6 Testig Hypotheses 6.1 Sir Roald Aylmer Fisher ( ) 1. (ull hypothesis) H 0 (alterative hypothesis) H T ( ), H 0,. 3. (sigificace level) 0 < α < 1 (critical regio)., H 0., 10%, 5%, 1%., T, T α (P (T W ) = α). ( H 1. ),. 4. T t, W (t W ). t W. T, H 0. α, H 0 (reject), H 1 (accept). t W. T, α, H 0 (, ) , 223.? 1. p. H 0 : p = 1 2 H 1 : p X. H 0, X B(400, 1/2) N(200, 10 2 )., Z = X 200 N(0, 1) 10.
26 26 6 Testig Hypotheses 3. α = 0.05., 5% ( ). 5% (= 2.5% ) 1.96, 4. x = 223 Z z = W : z = 2.3., H 0., 5% H 0.,. 5. 1%, 1% 2.58, z = % H 0.. α α α W W W W N(0, 1) α α z α ( ) m, σ 2, X = 1 ) X k N (m, σ2 X m σ/ N(0, 1) k=1,, (. N(m, σ 2 ) ). 6.2 ( ) 25 mm.,.,, 0.8 mm mm.? [ 5% H 0 : m = 25 ( ). 1%.]
27 (Two Types of Error) ( ) 120,., 25, , [ m. 5% H 0 : m = 25 ( ).] HW 18 ( ), [ 5% H 0 : p = 1/2 ( ). 1%..] HW 19 ( ), m = 60 (g).,, m 50 70, σ = 3 ( )., 25,, m = 60? [ 5% m = 60 ( ). 1%.] HW 20 ( ), 100g 2g., 2g. 200, 2.2g.,, 1.5g.. [ 5% ] (Two Types of Error) H 0, 4. \ H 0 H 0 H 0 2 H 0 1 α: 1 (Type I error) = β: 2 (Type II error) 1 = = 2 = = , 58.?
28 28 6 Testig Hypotheses. H 0 : p = 0.5 H 1 : p 0.5, α = B(100, 0.5) N(50, 5 2 ), B(100, 0.5). α p = , H 0,. 2., p, 2., p = 0.6. B(100, 0.6) N(60, 24) N(60, 5 2 ), B(100, 0.6) B(100, 0.5) 10..,, 2 β. β = 0.5. p = 0.50 p = 0.60 β (1) α β (2) α, β,. (3) H 0,. H 0.
29 29 7 Jerzy Neyma ( ) Ego Sharpe Pearso ( ) 7.1 ( ) m, σ 2,, X = 1 ) X k N (m, σ2 k=1 X m σ/ ( ).. N(0, 1) 7.2 ( : T - ) N(m, σ 2 ) X 1,..., X, U 2 = 1 (X i 1 X) 2,. X, i=1 T = X m U/ t 1 1 t (g) 9 494, 8 2.,? [ α = 0.05, t = 2.25 > H 0., N(0, 1), 2.25 < 1.96 H 0.] 7.2 ( ),. 50kg, 50kg. 12 (kg), x = 48.6, u 2 = [ 5% H 0 : m = 50 ( )]
30 30 7 HW A , A. A. [ 5% ] 7.3 P (P-value), α H 0.,, H 0. t, H 0, P = t, t P.,,. 7.3 A P. [0.0734] HW ,, ? P. [0.0076] ( ) X, Y a, b, E[aX + by ] = ae[x] + be[y ]. 7.5 ( ) X, Y a, b, V[aX + by ] = a 2 V[X] + b 2 E[Y ]. 7.6 ( ) 2 X N(m 1, σ1) 2 Y N(m 2, σ2) 2, a, b, ( ax + by N am 1 + bm 2, a 2 σ1 2 + b 2 σ2) 2
31 N(m 1, σ1), 2 N(m 2, σ2) 2 1, 2 X 1, X2, ( ) X 1 X 2 N m 1 m 2, σ2 1 + σ ( ). A B A 0.7, B [H 0 : m 1 = m 2, H 1 : m 1 m 2. z = % H 0.] HW 23 A 36, B 40, A x A = 64.5, B x B = A B., N(m 1, σ 2 ), N(m 2, σ 2 ) 1, 2 X 1, X2, U 2 1, U 2 2. U 2 = ( 1 1)U ( 2 1)U , T = X 1 X 2 ( ) U t ( ) 2 A, B. A 6, B 8. A : B : A,B. [ x A = , u 2 A = , x B = , u 2 B = , u 2 = , t = , t % %.]
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![33 8 Thomas Bayes (1702 1761) 8.1 Coditioal Probability 8.1 A, B 2. P (A) > 0, P (B A) = P (A B) P (A) A B. A, B. 8.2 (Drawig lots) 10, 2. 2 1,,? [,.] 8.3 2 X, Y ( X = Y ).](/docs-images/93/111303398/images/33-0.jpg "P (X 5 Y = 2) P (X + Y 8 X 4). [4/9, 5/9] HW 24 2 E, F, P (E) = 1 3, P (F ) = 1 2, P (E F ) = 1. 4. [2/3, 1/12, 1/4, 1/2, 1/6, 3/7] P (E c ), P (E F c ), P ((E F c ) c ), P (E F ), P (E F c ), P (E F E F ) 8.")
33 33 8 Thomas Bayes ( ) 8.1 Coditioal Probability 8.1 A, B 2. P (A) > 0, P (B A) = P (A B) P (A) A B. A, B. 8.2 (Drawig lots) 10, ,,? [,.] X, Y ( X = Y ). P (X 5 Y = 2) P (X + Y 8 X 4). [4/9, 5/9] HW 24 2 E, F, P (E) = 1 3, P (F ) = 1 2, P (E F ) = [2/3, 1/12, 1/4, 1/2, 1/6, 3/7] P (E c ), P (E F c ), P ((E F c ) c ), P (E F ), P (E F c ), P (E F E F ) 8.2 Idepedece of Evets A, B, P (A B) = P (A)P (B). A 1, A 2,..., A i1, A i2,..., A i (i 1 < i 2 < < i ). P (A i1 A i2 A i ) = P (A i1 )P (A i2 ) P (A i )
34 P (A) > 0, 2 A, B P (B) = P (B A) , 121, 211, , 1 1 A 1, 10 1 A 2, A 3. A 1, A 2, A 3 2, 3. HW 25 A, B, C, P (A) = a, P (B) = b, P (C) = c. a, b, c. [a(1 b), a + b ab, a + b + c ab bc ca + abc, a] P (A B c ), P (A B), P (A B C), P (A B C) 8.3 Bayes Formula 8.7 (Bayes formula) Ω = A 1 A 2, A 1 A 2 =, B, P (A 1 B) =. P (A 1 )P (B A 1 ) P (A 1 )P (B A 1 ) + P (A 2 )P (B A 2 ) 8.8 (1), A B, 95%, 2%... [0.160] (2), 100p %,. p? [1.9/( p)] HW 26, A B, 90%, 5%. (1). [ ] (2). [ ] HW 27 ( ) (1) 1, 6. [2/3] (2) 1, 6. [4/5]
35 X 1, X 2 [0, 1]., [0, 1]. X = (X 1 + X 2 )/2. a 0 < a < 1, A = ax 1 + (1 a)x 2. (1) E[A] = 1/2., A. (2) V[A] V[ X]., X A. 14 X 1, X 2 [0, 1]., [0, 1]. Y = X 1 X 2. E[Y ] = 4/9., Y ,?. 16 m, σ = 3, %, 95%. [14.5 ± 1.56, 14.5 ± 1.86] , 38, 62. [ 5% ] , 23.5 ( ) ?
36 cm cm, 4.63 cm. [ 1% ] 20 8%., 177, 23.. [ 5% 5% ] (kg), kg, [ x = 53.24, u 2 = 20.10, t = % ] 22, A B, 90%, 5%. (1).. [0.0674] (2).. [0.9938] 23, 100x % A (0 x 1). B, 90%, 5%.. x, x ( ) ,. 4. ( ),. 5.,.,.
37 37 9 χ 2 - Karl Pearso ( ) 9.1 χ 2-1 ( ) x 2 1 e x 2, x > 0, f (x) = 2 /2 Γ 2 0, x 0, 2 (χ 2 - ). (χ 2.), χ 2., Γ(t). = = = = = χ 2 - (1) X 1, X 2,..., X, N(0, 1), χ 2 = χ 2 -. (2) X 1, X 2,..., X, N(m, σ 2 ), χ 2 1 = 1 σ 2 i=1 i=1 X 2 i (X i X) 2, X = 1 i=1 X i ( ) 1 2. χ 2 1.
38 38 9 χ χ 2 - χ 2, m =, σ 2 = (Goodess of Fit Test) A 1, A 2,..., A k k., X 1, X 2,..., X k. A 1 A 2 A k p 1 p 2 p k 1 X 1 X 2 X k, p 1, p 2,..., p k. 9.2 (Pearso χ 2 - ) m i = p i, χ 2 k 1 = k (X i m i ) 2 m i=1 i, m 1,..., m k (m i = p i 5), k , 120.? [χ 2 = 2.9. χ 2 5-5% %.] 9.4,, 1 1 (2013 J ) , 1.436, λ = (i) m i = p i 5 0, 1,..., 5 6. (ii), = 4 2.
39 HW 28,. 4 : 3 : 2 : 1.,? [χ 2 = χ 2 3(0.05) = ] A O B AB HW 29, 45, 55.? (1) (2), A = {A 1,..., A r }, B = {B 1,..., B s }, χ 2 = r i=1 s j=1 ( Xij X i X i X j ) 2 X j, (X ij 5), (r 1)(s 1) 2. B 1 B 2 B s A 1 X 11 X 12 X 1s X 1 A 2 X 21 X 22 X 2s X 2. A r X r1 X r2 X rs X r X 1 X 2 X s.
40 40 9 χ ? [χ 2 = %, χ 2 1(0.01) = ] , , 5., 5 1:1?. [χ 2 = χ 2 5(0.01) = :1. 51:49, χ 2 = 7.97, 5.] : 0:5 1:4 2:3 3:2 4:1 5: , 1 ( ) , λ = ?
41 : P (χ 2 χ 2 (α)) = α α χ α \α ( = 1 ).
42
43 43 第 10 章 多変量の統計 Sir Fracis Galto ( ) 変量データの記述 2 変量データ (2 次元データ): (x1, y1 ), (x2, y2 ),..., (x, y ) scatter diagram (散布図) データを xy-座標平面に図示したもの 例 題 10.1 身長 (x) と体重 (y) の散布図. クラス (A) とクラス (B) に対する結果 (B) (A) covariace (共分散) 個の 2 変量データ (x1, y1 ), (x2, y2 ),..., (x, y ) に対して, 変数ご との平均値と分散 1 x = xi, i=1 1 = (xi x )2 ; i=1 σx2 1 y = yi, i=1 1 = (yi y )2 i=1 σy2 を用いて共分散が定義される: 1 1 = (xi x )(yi y ) = xi yi x y i=1 i=1 σxy (注意) σxy = σyx. σxx = σx2 (したがって, 分散を σxx と書く流儀もある). correlatio coefficiets (相関係数) r = rxy = (注意) rxy = ryx. 正の相関 負の相関 強い相関 弱い相関 無相関 σxy σxy = σx σy σxx σyy
44 ( ( )) x i = x i x σ x, ỹ i = y i ȳ σ y x, y, x, ỹ, r xy = σ xỹ = r xỹ (10.1)., x, y, x, ỹ r xy 1. {t(x i x) + (y i ȳ)} 2 0 t (A) (B) A B HW 30 2 (x 1, y 1 ), (x 2, y 2 ),..., (x, y ) σ x > 0, σ y > 0., r = 1., r = 1. [ 10.4.] 10.2 Radom Vectors X, Y, covariace ( ) σ XY = Cov (X, Y ) = E[(X E[X])(Y E[Y ])] = E[XY ] E[X]E[Y ]., correlatio coefficiet ( ) : r XY = σ XY σ = XY σ X σ Y σxx σy Y
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46 Q a = Q b = 0, 1, a 0 = σ xy σ 2 x y = a 0 x + b 0., b 0 = ȳ a 0 x (10.2) (x 1, y 1 ), (x 2, y 2 ),..., (x, y ), x, y y ȳ = σ xy (x x) = σ y r(x x) σx 2 σ x y ȳ σ y., y, x x x = σ xy (y ȳ) = σ x r(y ȳ) σy 2 σ y., r. x x σ x = r x x σ x (10.3) = r y ȳ σ y (10.4) ( ) 2, ( x, ȳ), ( ) A,B (x) (y). A, x = , ȳ = 63.59, σ 2 x = , σ 2 y = , σ xy = , x,., y y = 0.73x (10.5) x = 0.27y (10.6). (10.6) 1/ , (10.5)., B,, x,, y. x = , ȳ = 51.05, σ 2 X = , σ 2 Y = , σ XY = y = 0.72x x = 0.58y
47 10.3. Regressio Models (A) (B) HW 32 4 (0, 1), (1, 3), (3, 6), (4, 6) x. [y 4 = 1.3(x 2)] 27 x, y, σ xy σ x σ y. 28 Galto, (1886).. Galto.,? ( ). Mid-height parets (x) Adult Childre (y) below above sum above below sum
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II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =
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数学の基礎訓練I
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第７章
5. 推定と検定母集団分布の母数を推定する方法と仮説検定の方法を解説する まず 母数を一つの値で推定する点推定について 推定精度としての標準誤差を説明する また 母数が区間に存在することを推定する信頼区間も取り扱う 後半は統計的仮説検定について述べる 検定法の基本的な考え方と正規分布および二項確率についての検定法を解説する 5.1. 点推定先に述べた統計量は対応する母数の推定値である このように母数を一つの値およびベクトルで推定する場合を点推定
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最小2乗法
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Excelにおける回帰分析（最小二乗法）の手順と出力
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