1
1 1.1 1.1.1 1 168 75 2 170 65 3 156 50... x y 1 x 1 y 1 2 x 2 y 2 3 x 3 y 3... x ( ) 2
1 1 0 1 0 0 2 1 0 0 1 0 3 0 1 0 0 1...... 1.1.2 x = 1 n x (average, mean) x i s 2 x = 1 n (x i x) 2 3
x (variance) s x = 1 (x i x) n 2 x 1.1.1 ( ). λ > 1 λ x i x λs x x i n λ 2 λs x n λ 2 ( ) x i x > λs x x i m ns 2 x = (x i x) 2 (x i x) 2 mλ 2 s 2 x x i x λs x 1.1.2. x, y a, b y = ax + b ȳ = a x + b, s 2 y = a 2 s 2 x ( ) ȳ = a x + b ȳ = 1 n = 1 n ax i + b ax i + 1 n = a x + b b 4
s 2 y = a 2 s 2 x s 2 y = 1 n = 1 n = 1 n (y i ȳ) 2 (ax i + b a x b) 2 a 2 (x i x) 2 = a 2 s 2 x x x s x z = x x s x z = 0, s z = 1 SS = 10z + 50 SS = 50, s SS = 10 z, SS 1.2 1.2.1 data1 demo.xls 5
1. ( ) 2. ( ) 3. 1.2.2 s xy = 1 (x i x)(y i ȳ) n x, y (covariance) p i = (x i x)(y i ȳ) 1. ( ) ( ) p i ( x, ȳ) ( x, ȳ) A p i B y B A B B (x,y) x A B 2. ( ) ( ) p i A p i B 6
y A B B (x,y) B x B A 3. p i p i 1. s xy 2. s xy 3. s xy ( ) 1.2.3 r = s xy s x s y x, y (correlation coefficient) B s x, s y x i x, y i ȳ 7
r 1 r = 1 x a b c y d e f x = ȳ = 0 r 2 = (ad + be + cf) 2 (a 2 + b 2 + c 2 )(d 2 + e 2 + f 2 ) 1 (ae bd) 2 + (bf ce) 2 + (cd af) 2 0 d a = e b = f c ( ) a = 0 d = 0 or b = c = 0 (a, d) y 1. r 1 2. r 1 3. r X = (x 1,..., x n ), Y = (y 1,..., y n ) 8
θ cos θ ( ) ( ) 1.3 1.3.1 x, y (x 1, y 1 ),..., (x n, y n ) y x y = f(x) x y f(x) 1. f(x) f(x) = a + bx, f(x) = ax b, f(x) = a + bx + cx 2 2. b 0,..., b p a, b a, b, c S S = (y i f(x i )) 2 S S y = f(x) y 2 y = f(x) y = f(x) 9
y y=f(x) x 3. b 0,..., b p S b i = 0 (i = 0,..., p) S f(x) f(x) ( ) (regression) (x 1,..., x p, y) (x 11,..., x p1, y 1 ),..., (x n1,..., x pn, y n ) y x 1,..., x n y = b 0 + b 1 x 1 + + b p x p S = (y i (b 0 + b 1 x 1i +... + b p x pi )) 2 b 0, b 1,..., b p S b i = 0 (i = 0,..., p) (1) 10
(1) b 0,..., b p b 0 + b 1 x 1 + + b p x p p = 1 f(x 1,..., x p ) (1) b 0,..., b p ( ) 2 y ȳ = s xy s 2 (x x) = r s y (x x) x s x x, y ( ) ( R ( ) ) 1.3.2 R R 2 = 1 (y i f(x i )) 2 (y i ȳ) 2 y = f(x) f { } 2 (x i x)(y i ȳ) (x i x) 2 n (y i ȳ) 2 R2 y ȳ = s xy s 2 x x x 11
( ) 2 x, y ( ) 1.3.1. 2 1.3.2. BW = f(sl) R 0.978 0.982 R BW SL BW = asl b SL BW data1 demo.xls R (y i f(x 1i,..., x pi )) 2 R 2 = 1 (y i ȳ) 2 y y i f(x 1,..., x p ) 12
1.4 1. data1.xls ( web ) 2. katakuchi.csv 3. buri.csv 4. questionnaire.csv 5. data1.xls Excel katakuchi.csv, questionnaire.csv R Excel 1. data1.xls Excel 2. Excel average( ) varp( ) stdevp( ) max( ) min( ) correl( 1, 2) Excel Excel Excel C4 =B4 C4 B4 C4 C5 C5 B4 B5 C4 =B4 B4 C4 C5 B5 13
B4 $B$4 $ B4 F4 $B4 B$4 3. frequency 8 9 frequency(, ) shift, ctrl Enter frequency shift, ctrl Enter 4. 5. 6. 0.5 2.0 0.5 8.0 R- R 1. R 14
2. katakuchi.csv 3. Enter q() R d<-read.table("katakuchi.csv",header=t) min(d$sl) max(d$sl) hist(d$bw,breaks=seq(4,29)) lm(bw~sl,d) d<-d[order(d$sl),,] fm<-nls(d$bw~a*d$sl ^b,d,start=list(a=0.1,b=1)) p<-predict(fm) plot(d$sl,d$bw) par(new=t) plot(d$sl,p,type="l") q() 1. 2. (q) (p) p a q b = c a, b, c a log p + b log q = log c 15
log p log q Excel 1. data1.xls 2. 3. 2 4. 5. R 1. R 2. questionnaire.csv 3. Enter q() R d2<-read.table("questionnaire.csv",header=t,sep= \t ) table(d2$p1,d2$p2) q() 16
2 2.1 2.1.1 A P (A) A 0 P (A) 1 P (A) A P (A) = 0 A P (A) = 1 A 2.1.1. 1... 6 1/6 A B A B A B A B A B P (A B ) = P (A)+P (B) P (A B ) = P (A) + P (B) P (A, B ) A B P (A B ) = P (A)P (B) 17
( ) P (A, B ) = P (A)P (B A) P (B A) A B 2.1.2. 10 7 5 3 8 3 3 8 8 P ( ) = 3 8 (1) 17 P ( ) = 3/17 8/17 = P ( ) = 3 P ( ) 8 P ( ) = P ( P ( ) (1) P (A B) = P (A B ) P (B ) 18
2.1.1. P (A B) = P (A B) P (B) P (B) = 0 2.1.1. P (A B) = P (B)P (A B) 2.1.2. A B P (A B) = P (A)P (B) P (A B) = P (B)P (A B) P (A B) = P (A) P (A B) = P (A)P (B A) P (B A) = P (B) A 1, A 2,..., A n p(a 1 A 2... A n ) = P (A 1 )P (A 2 ) P (A n ) 2.1.3. A B A, B P (A) + P (B) = P (A or B) = 1 P (A) = 1 2, P (B) = 1 2 19
2.1.4. A, B p 1 p n A k p k (1 p) n k A k n A k ( ) n n! = k k!(n k)! ( ) n P (A k )) = p k (1 p) n k k 2.1.2 X k P (X = k) x 1, x 2, x 3,... X 1,..., X n x 1,..., x n X 1 = x 1,..., X n = x n X 1,..., X n 2.1.5. X X n Y Y, Z = Y n 20
Y B(n, p) X 1,..., X n X Y = X 1 + + X n, Z = X 1 + + X n n 2 p = 1 2 p 1 2 n n X X x 1, x 2... E(X) = i x i P (X = x i ) X X V (X) = E((X E(X)) 2 ) X X, Y X, Y C(X, Y ) = E((X E(X))(Y E(Y ))) 2.1.6. n v 1,..., v n v i f i N = n f i X P (X = v i ) = f i N X E(X) = v i f i N 2.1.1.1. X E(aX + b) = ae(x) + b, X, Y E(X + Y ) = E(X) + E(Y ) V (ax + b) = a 2 V (X) 21
( ) p ij = P (X = x i, Y = y j ) E(X + Y ) = i,j (x i + y j )p ij = i,j x i p ij + i,j y j p i,j = i x i j p ij + j y j ( i p i,j ) = i x i P (X = x i ) + j y j P (Y = y j ) = E(X) + E(Y ) 2.1.2. E(a 1 X 1 + + a n X n + b) = a 1 E(X 1 ) + + a n E(X n ) + b ( ) 2.1.2.1. X, Y V (X + Y ) = V (X) + V (Y ) + 2C(X, Y ), C(X + Y, Z) = C(X, Y ) + C(Y, Z) C(X, Y ) = E(XY ) E(X)E(Y ) X, Y E(XY ) = E(X)E(Y ) C(X, Y ) = 0 22
( ) V (X + Y ) = E({X + Y E(X + Y )} 2 ) = E({(X E(X)) + (Y E(Y ))} 2 ) = E((X E(X)) 2 ) + 2E((X E(X))(Y E(Y ))) + E((Y E(Y )) 2 ) = V (X) + V (Y ) + 2C(X, Y ) C(X, Y ) = E((X E(X))(Y E(Y ))) = E(XY XE(Y ) Y E(X) + E(X)E(X)) = E(XY ) E(X)E(Y ) E(X)E(Y ) + E(X)E(Y ) = E(XY ) E(X)E(Y ) X, Y x, y x i, y j P (X = x i ) = p i, P (Y = y j ) = q j E(XY ) = x i y j p i q j i,j = x i p i y j q j i j = E(X)E(Y ) 2.1.3. ( n ) V a i X i + b = a 2 i V (X i ) + 2 i<j X 1,..., X n ( n ) V a i X i + b = a 2 i V (X i ) a i a j C(X i, X j ) ( ) 2.1.7. 2 2 Y = X 1 + + X n Z = Y P (X = 1) = p n E(X) = p, V (X) = p(1 p) E(Y ) = np V (Y ) = np(1 p), E(Z) = p V (Z) = 23 p(1 p) n
2.1.8. µ µ µx P (X = x) = e, x = 0, 1, 2,... x! x=0 µ x x! = eµ X E(X) = µ, V (X) = µ ( ) E(X) = µ µ µx E(X(X 1)) = x(x 1)e x! = e µ µ 2 µ x 2 (x 2)! = µ2 x=2 E(X 2 ) = E(X(X 1)) + E(X) = µ 2 + µ V (X) = E(X 2 ) E(X) 2 = µ 2 + µ µ 2 = µ B(n, p) np = µ n = P (X = x) = n C x p x (1 p) n x n! ( µ ) x ( 1 µ ) n x x!(n x)! n n ( 1 1 ) ( 1 x 1 n n = µx x! µx x! e µ (n ) ) { ( 1 µ n x=2 ) n } µ(1 x n) µ (p) (n) µ = np X X 24
2.1.3 X n X 1,..., X n n X X X X X X X X 1, X 2 X = X 1+X 2 2 X = 0.5 X = 0.5 (X 1, X 2 ) = (1, 0), (0, 1) X = 0, 1 (X 1, X 2 ) = (0, 0) (X 1, X 2 ) = (1, 1) 0.5 (x + y) n 2.1.4 ( ). X X X X E(X) ϵ > 0, V (X) < K V ( X) < K n, E( X) = E(X) V ( X) K n > V ( X) x E(X) ϵ ϵ 2 x E(X) ϵ (x E(X)) 2 P ( X = x) P ( X = x) = ϵ 2 P ( X E(X) ϵ) P ( X E(X) ϵ) K nϵ 2 25
1. A p n A p ( ) A X X 1 p + 0 (1 p) = p X A A p 2. X X k X = k A ( ) X A A { 1 X = k X(k) = 0 X k X 1 (k),, X n (k) X = k S n,k = X i (k) S n,k X(k) n P (X = n k) np (X = k) 26
2.1.4 octave octave X 0.5 F X F (x) = P (X x) Y [0, 1] X = F 1 (Y ) X F 1 (y) F (x) = y x octave 2.2 2.2.1 2.2.1. [a, b] 27
X X 1 p(x) P (a X b) = b a p(x)dx ( ) p(x) p(x)dx = 1 ( ). F (x) = P (X x) E(X) E(X) = xp(x)dx V (X), C(X, Y ) V (X) = E((X E(X)) 2 ) C(X, Y ) = E((X E(X))(Y E(Y ))) 2.2.2. p(x) = 1 (x [0, 1]), p(x) = 0 (x / [0, 1]) E(X) = 1 0 xdx = 1 2 V (X) = 1 0 ( x 1 ) 2 dx = 1 2 12 28
2.2.3. p(x) = 1 e (x µ)2 2σ 2 2πσ µ σ N(µ, σ 2 ) X N(µ, σ 2 ) µ X σ 2 X ( ) ( ) z = x µ σ e x2 dx = π E(X) = 1 2πσ xe (x µ) 2 2σ 2 dx E(X) = µ V (X) = 1 2πσ V (X) = 1 2π (x µ) 2 e (x µ) 2 2σ 2 dx σ 2 z 2 e z2 2 dz z( ze z2 2 ) V (X) = σ 2 N(µ, σ 2 ) p(x) x = µ x = µ x = µ σ, x = µ+σ / / ( ) 29
X, Y X, Y [a, b], [c, d] f(x, y) P (a X b, c Y d) = d b c a f(x, y)dxdy f(x, y) (X, Y ) P (a X b, < Y < ) = P ( < X <, c X d) = f(x, y)dy, b a d c f(x, y)dx f(x, y)dydx f(x, y)dxdy X, Y X, Y [a, b], [c, d] a x b, c y d X, Y p(x), q(y) X, Y f(x, y) X, Y X, Y f(x, y) = p(x)q(y) ( ) P (a X b, c Y d) = P (a X b)p (c Y d) = = b p(x)dx a d b c a d c q(y)dy p(x)q(y)dxdy 2.2.1. X, Y p(x), q(y) (X, Y ) f(x, y) E(φ(X, Y )) = φ(x, y)f(x, y)dxdy 30
( ) Z = φ(x, Y ) r(z) φ(x, Y ) 0 E(Z) = = = 0 t tr(t)dt 0 0 0 z r(t)dzdt r(t)dtdz {(t, z) : 0 t <, 0 z t} = {(t, z) : 0 z <, z t < } E(Z) = = = = = 0 z 0 r(t)dtdz P (Z > z)dz 0 {(x,y): φ(x,y)>z} φ(x,y) 0 f(x, y)dxdydz f(x, y)dzdxdy φ(x, y)f(x, y)dxdy {(x, y, z) : 0 z <, φ(x, y) > z} = {(x, y, z) : x, y <, 0 z < φ(x, y)} 2.2.1.1. X E(aX + b) = ae(x) + b, V (ax + b) = a 2 V (X) X, Y E(X + Y ) = E(X) + E(Y ) ( ) X, Y 31
f(x, y) E(X + Y ) = = = (x + y)f(x, y)dxdy xf(x, y)dxdy + ( ) x f(x, y)dy dx + = E(X) + E(Y ) yf(x, y)dxdy ( y f(x, y)dx ) dy 2.2.2. E(a 1 X 1 + + a n X n + b) = a 1 E(X 1 ) + + a n E(X n ) + b 2.2.2.1. X, Y V (X + Y ) = V (X) + V (Y ) + 2C(X, Y ), C(X + Y, Z) = C(X, Y ) + C(Y, Z) C(X, Y ) = E(XY ) E(X)E(Y ) X, Y E(XY ) = E(X)E(Y ) C(X, Y ) = 0 ( ) X, Y X, Y p(x), q(y) f(x, y) E(XY ) = = = = xyf(x, y)dxdy xyp(x)q(y)dxdy yq(y) xp(x)dx = E(X)E(Y ) xp(x)dxdy yq(y)dy 32
2.2.3. V ( a i X i + b) = a i a j C(X i )C(X j ) a 2 i V (X i ) + 2 i<j X 1,..., X n V ( a i X i + b) = a 2 i V (X i ) ( ) 2.2.4. X 1,..., X n X i N(µ i, σ 2 i ) a i ( n a i X i N a i µ i, ) a 2 i σi 2 X i N(µ, σ 2 ) X 1 + + X n N n ( µ, σ 2 ) n 2.2.5 ( ). X µ σ 2 n X X N(µ, σ2 n ) X X 1 + + X n N(nµ, nσ 2 ) 33
2.2.4. X A 1 0 P (X = 1) = p X n Y B(n, p) E(X) = p, V (X) = p(1 p) B(n, p) n ( ) p(1 p) N(np, np(1 p)) X N p, n 2.2.2 X ( ) 2.2.1. χ 2 ( ) n n χ 2 1 ) n f n (χ 2 ) = 2 n 2 Γ( n 2 )(χ2 2 1 e 1 2 χ2 (χ 2 > 0) 0 (χ 2 0) Γ Γ(p) = 0 x p 1 e x dx, (p > 0) 2.2.6. Z N(0, 1) Z 2 1 χ 2 ( ) 34
2.2.7. X 1,..., X n N(µ, σ 2 ) 1 σ 2 (X i µ) 2 n χ 2 ( ) 2.2.8. X 1,..., X n N(µ, σ 2 ) 1 σ 2 (X i X) 2 n 1 χ 2 ( ) 2.2.2. t n n t 1 f n (t) = ( nb n 2, ) 1 2 ( ) n+1 1 + t2 2 n B(p, q) B(p, q) = 1 t 0 x p 1 (1 x) q 1 dx, (p > 0, q > 0) 2.2.9. Z N(0, 1) χ 2 n χ 2 n t ( ) t = Z χ 2 n 2.2.10. X 1,..., X n N(µ, σ 2 ) U 2 = 1 n 1 (X i X) 2 35
t = X µ U 2 n n 1 t U 2 ( ) 2.2.11. 2 N(µ X, σ 2 ), N(µ Y, σ 2 ) X, Y m, n X, Ȳ U 2 X, U 2 Y T = X Ȳ (µ X µ Y ) V ( 2 1 m + ), V 2 = (m 1)U X 2 + (n 1)U Y 2 1 m + n 2 n T m + n 2 t 2.2.3. F m, n (m, n) F m m 2 n n 2 F m 1 2 B( f (m,n) (F ) = m 2, n 2 ) (F > 0) (mf +n) m+n 2 0 (F 0) B(p, q) 2.2.12. χ 2 1, χ 2 2 m, n χ2 (m, n) F ( ) F = χ 2 1 m χ 2 2 n 2.2.3 octave 36
X i, i = 1,... n [0, 1] µ = E(X i ) = 0.5 σ 2 = V (X) = 1 12 n = 50 X = X 1 + + X n n N(0.5, 1/600) octave 0.01 1000 Z 1000/100=10 octave:1> for :1000 > data(i)=mean(rand(50,1)); > end octave:2> x=0:0.01:1; octave:3> hist(data,x, facecolor, w, edgecolor, b ) octave:4> hold octave:5> plot(x,10*normpdf(x,0.5,sqrt(1/600)), color, r ) 37
3-3.1 sample population X 1,..., X n X i X i X i random sampling statistical inference 1 parametric method estimation test 2 yes,no 3.1.1. 1 inference 38
1. 2. 1 2 ( ) 2 3.2 3.2.1 ( parameter) µ X ( ) E( X) = µ 3.2.1. 1. 2. ( ) ( ) U 2 = 1 n 1 (X i X) 2 39
( ) µ σ 2 ( ) E( X) X1 + + X n =E n = 1 n E(X 1 + + X n ) = 1 n {E(X 1) + + E(X n )} = E(X) = µ (X i µ) 2 = n( X µ) 2 + (X i X) 2 (X i µ) 2 = (X i X + X µ) 2 = (X i X) 2 + 2( X µ) (X i X) + n( X µ) 2 = n( X µ) 2 + (X i X) 2 1 0 E((X i µ) 2 ) = ne ( ( n ) ( X µ) 2) + E (X i X) 2 nσ 2 nσ 2 = ne ( ( n ) ( X µ) 2) + E (X i X) 2 1 E( X) = µ, V ( X) = σ2 nσ 2 = n σ2 n + E n ( n ) (X i X) 2 40
( ) 1 E (X i n 1 X) 2 = σ 2 n X 1,..., X n θ T n ε > 0 lim n P ( T n θ > ε) = 0 T n θ 3.2.2 3.2.1. 1 A B C 1 3, 1 2, 2 3 80 49 31 A B C most likely A B C 49 31 A ( ) (1 ) 49 ( 80 1 1 31 = 0.000 49 3 3) B ( ) (1 ) 49 ( 80 1 1 31 = 0.012 49 2 2) 41
C ( ) (2 ) 49 ( 80 1 2 31 = 0.054 49 3 3) 49 31 C C 2/3 likelihood X 1,..., X n f(x µ, σ) = 1 e (x µ)2 2σ 2 2πσ x 1,..., x n f(x 1,..., x n µ, σ) = n 1 e (x i µ) 2σ 2 2πσ L = log f(x 1,..., x n µ, σ) 1 = log e (x i µ) 2σ 2 2πσ =n log 1 2πσ 2 (x i µ) 2 = n log( 2πσ) 1 2σ 2 2σ 2 2 (x i µ) 2 42
L µ, σ 2 µ, σ 2 L µ = 0, L σ = 0 ˆµ = 1 n x i ˆσ2 = 1 n (x i ˆµ) 2 L ˆσ 2 (x 1, y 1 ),..., (x n, y n ) y = f(x) x y f(x) f(y f(x), σ) = 1 e (y f(x))2 2σ 2 2πσ f(y 1,..., y n f(x), σ) = n 1 e (y i f(x i )) 2σ 2 2πσ f L = n log( 2πσ) 1 2σ 2 (y i f(x i )) 2 (y i f(x i )) 2 f f 2 2 97 1 2 43
3.2.3 P (X D) = P (X D) P (D) P (X D) = P (X D)P (D) X, D P (D X) = P (D X)P (X) P (X D)P (D) = P (D X)P (X) P (X D) = P (D X)P (X) P (D) 3.2.2. 2 A B C 1 3, 1 2, 2 3 80 49 31 A B C A B C 49 31 0.000, 0.012, 0.054 49 31 C C A 3 B 6 C 1 10 1 B D 49 31 A A P (A D) = P (D A)P (A) P (D) 44
P (D A) = 0.000 P (A) = 0.3 D A, D B, D C D P (D) = P (D A)P (A)+ P (D B)P (B) + P (D C)P (C) = 0.0126 A P (A D) = 0.000 0.3 0.0126 B B P (B D) = 0.012 0.6 0.0126 = 0 = 0.571 C C P (C D) = 0.054 0.1 0.0126 = 0.429 B B A B C 0.000, 0.571, 0.429 3.2.3. PC X P (X = ) P (X = ) 1000 P (X = ) = 0.348 P (X = ) = 0.652 Y 1, Y 2 { 1 Y 1 = 0 45
Y 2 = { 1 0 P (Y 1 = 1 X = ) = P (Y 1 = 1 X = ) P (X = ) P (Y 2 = 1 X = ) = P (Y 2 = 1 X = ) P (X = ) P (Y 1 = 1 X = ) = P (Y 1 = 1 X = ) P (X = ) P (Y 2 = 1 X = ) = P (Y 2 = 1 X = ) P (X = ) = 0.271 = 0.172 = 0.136 = 0.523 P (X = Y 1 = 0 Y 2 = 1) P (X = Y 1 = 0 Y 2 = 1) X X, Y 1, Y 2, P (X Y 1, Y 2 ) = P (Y 1, Y 2 X)P (X) P (Y 1, Y 2 ) X P (Y 1, Y 2 X) = P (Y 1 X)P (Y 2 X) 46
P (Y 1, Y 2 X)P (X) P (Y 1 = 0, Y 2 = 1 X = )P (X = ) =P (Y 1 = 0 X = )P (Y 2 = 1 X = )P (X = ) =(1 0.271) 0.172 0.348 = 0.044 P (Y 1 = 0, Y 2 = 1 X = )P (X = ) =P (Y 1 = 0 X = )P (Y 2 = 1 X = )P (X = ) =(1 0.136) 0.523 0.652 = 0.294 P (Y 1, Y 2 ) P (Y 1, Y 2 ) =P (Y 1, Y 2 X = )P (X = ) +P (Y 1, Y 2 X = )P (X = ) =0.338 X P (Y 1, Y 2 X) P (X = Y 1 = 0 Y 2 = 1) = 0.044/0.338 = 0.129 P (X = Y 1 = 0 Y 2 = 1) = 0.294/0.338 = 0.871 2 http://gihyo.jp/dev/serial/01/machine-learning 47
3.3 θ α 0.05 a < θ < b 1 α [a, b] α = 0.05 1 α = 0.95 = 95 β = 1 α [a, b] 100(1 α) a, b α Z Z 100(1 α) = 100β θ θ θ 0 θ θ 0 θ 1. β = 1 α [a, b] θ 0 θ 0 α 2. θ [a, b] (1) θ (2) θ θ 0 P ( θ θ 0 b a) 1 α 3.3.1 ( ) n X 1,..., X n X µ σ 2 N(µ, σ 2 ) ( ) 2.2.4 X N(µ, σ 2 /n) (n > 100) 2.2.5 X i X N(µ, σ 2 /n) 48
X µ N(0, σ 2 /n) σ 2 P ( X µ > z(α/2)) = α/2 z(α/2) > 0 P ( X µ < z(α/2)) = α/2 P ( z(α/2) X µ z(α/2)) = 1 α P ( X z(α/2) µ X z(α/2)) = 1 α ( ) [ X z(α/2), X +z(α/2)] 100(1 α)% z(α/2) N(0, σ 2 /n) α 100α α/2 α/2 -z(α/2) z(α/2) ( ) N(0, 1) N(0, 1) z(α) ( ) N(µ, σ 2 ) z(α) 1. n 2. σ 2 49
3. β α = 1 β 4. X 5. σ 2 n N(0, σ 2 /n) α z(α/2) 6. [ X z(α/2), X + z(α/2)] 7. µ 0 µ 0 1 α [ X z(α/2), X + z(α/2)] α µ µ 0 µ µ 0 ( ) ( ) t ( ) N(µ, σ 2 ) ( ) n X 1,..., X n X U t = X µ U n n 1 t ( t ) P ( t > t n 1 (α/2)) = α ( ) t n 1 (α) 100(1 α)% µ X t n 1(α/2)U n µ X + t n 1(α/2)U n 50
t 1. 2. β α = 1 β 3. X 4. σ 2 n n 1 t α t n 1 (α/2) 5. [ X t n 1 (α/2) U 2 n, X + tn 1 (α/2) U 2 n ] 6. µ 0 µ 0 1 α [ X t U n 1 (α/2) n, X+ U t n 1 (α/2) n ] α µ µ 0 t ( ) ( ) F = t 2 (1, n 1) F 3.3.1. t 20 125.8 1. σ = 4.57 2. σ = 4.57 95% 99% 51
µ µ 0 µ 0 µ 0 < X z(α/2) X + z(α/2) < µ 0 X µ 0 < z(α/2) z(α/2) < X µ 0 t µ µ 0 µ 0 µ 0 < X U t n 1 (α/2) 2 n X U +t n 1 (α/2) 2 n < µ 0 3.3.2 ( ) ( ) p X 1 A X = 0 E(X) = p 1 + (1 p) 0 = p p 1 α ( ) Z n nz p F p 52
Z = X 1 + + X n X i p p(1 p) = n p(1 p) ( 2 + (1 p)(0) p) 2 2.2.5 p(1 p) N p, Z p ( n ) p(1 p) N 0, n P (Z p < z(α/2)) = P (z(α/2) < Z p) = α 2 z(α/2)( 100α ) z(α/2) < Z p < z(α/2) Z z(α/2) < p < Z + z(α/2) 100(1 α) [Z z(α/2), Z + z(α/2)] 100(1 α) ( ) p(1 p) N 0, p n z(α/2) p Z p(1 p) ( )σ = Z p N(0, 1) n σ p Z z 0 (α)σ < p < Z + z 0 (α)σ z 0 (α) P (Z p < z 0 (α)) = α 2 p σ Z z 0 (α)σ < p < Z + z 0 (α)σ p p (n + z o (α) 2 )p 2 (z 0 (α) 2 + 2npZ)p + nz 2 < 0 z 2 + 2nZ (z 0 (α) 2 + 2nZ) 2 4(n + z 0 (α) 2 )nz 2 2(n + z 0 (α) 2 ) < p < z 2 + 2nZ + (z 0 (α) 2 + 2nZ) 2 4(n + z 0 (α) 2 )nz 2 2(n + z 0 (α) 2 ) n (n + z o (α) 2 )p 2 (z 0 (α) 2 + 2npZ)p + nz 2 = n(p 2 2pZ + Z 2 ) + z 0 (α) 2 p(1 p) n(p 2 2pZ + Z 2 ) + z 0 (α) 2 Z(1 Z) Z(1 Z) Z(1 Z) Z z 0 (α) < p < Z + z 0 (α) n n 53
Z(1 Z) z 0 (α) N n ( 0, ) Z(1 Z) α z(α/2) n Z z(α/2) < p < Z + z(α/2) ( ) p(1 p) Z p N 0, ( ) n Z(1 Z) N 0, n ( ) X X t ( ) p p(1 p) ( Z = 0 [0, 1 α 1/n ] Z = 1 [α 1/n, 1] F ( ) nz B(n, p) x nz P (nz x) = P (F > f) F (2(n x + 1), 2x) F x(1 p) f = (n x + 1)p F P (F > f 1 ) = α/2 x(1 p) f 1 f 1 = (n x + 1)p p p 1 p 1 = x (n x + 1)f 1 + x nz B(n, p 1 ) P (nz x) = α/2 P (nz x) = P (F > f ) F (2(x + 1), 2(n x)) F f (n x)p = (x + 1)(1 p) 54
P (F > f 2 ) = α/2 f 2 f 2 = (n x)p (x + 1)(1 p) p p 2 p 2 = (x + 1)f 2 (x + 1)f 2 + (n x) nz B(n, p 2 ) P (nz x) = α/2 p 100(1 α)% [p 1, p 2 ] 3.3.2. 20 0.35 95% 3.3.3 S 2 χ 2 = ns2 n 1 100(1 α) σ 2 ns ( 2 χ 2 α ) < σ 2 ns < ( 2 ) n 1 2 χ 2 n 1 1 α 2 3.4 3.4.1. 1000 1 1 55
1 1 1 6 1 ( ) 1 1 6 = 0.166666667 4 1 ( 1 4 6) = 0.000771605 4 1 1000 4 1 0 1000 1 0.0771605... P P α ( )α 1 1 1 6 1 6 1 4 1 4 56
6 1 P P 6 3.4.1 ( ) µ [ ] µ 0 µ = µ 0 H 0 : µ = µ 0 X µ 0 N ( 0, σ 2 /n ) N ( 0, σ 2 /n ) P (z > z(α/2)) = α/2 z(α/2) α X µ 0 < z(α/2), z(α/2) < X µ 0 X µ 0 > z(α/2) X µ 0 X µ 0 z(α/2) N(0, σ 2 /n) α 100α H 0 : µ = µ 0 µ = µ 0 57
α/2 α/2 -z(α/2) z(α/2) P P X µ 0 ζ X µ 0 = ζ X µ 0 ζ X > µ 0 X µ 0 ζ P P p p -ζ ζ 1. n 2. σ 2 58
3. µ µ 0 µ µ 0 4. α 5. X µ 0 6. σ 2 n N(0, σ 2 /n) α z(α/2) 7. X µ 0 > z(α/2) µ µ 0 P 4 5. X µ 0 6. σ 2 n N(0, σ 2 /n) Z P ( X µ 0 < Z) = p 7. p < α α 2α µ µ 0 [ ] µ µ 0 µ > µ 0 µ µ 0 µ µ 0 H 0 : µ = µ 0 X µ 0 N ( 0, σ 2) N ( 0, σ 2) P (t > z(α)) = α z(α) α z(α) < X µ 0 X µ 0 X µ z(α) N(0, σ 2 /n) α 59
α z(α) H 0 : µ = µ 0 µ 1 < µ 0 µ 1 H 0 : µ = µ 1 X (µ = µ 0 ) µ µ = µ 1 µ = µ 1 X µ 1 N ( 0, σ 2) X µ 0 < X µ 1 X µ 1 t ( ) H 0 : µ = µ 0 N(µ, σ 2 ) n X 1,..., X n X U t = X µ U n n 1 t U 2 t 60
P (t > t n 1 (α)) = α/2 t n 1 (α) α t > t n 1 (α) 3.4.2. X µ µ 0 17 ν 0 1. µ µ 0? 2. µ > µ 0? 3. µ < ν 0? (1 ) H 0 : µ = µ 0 (2 ) ( ) H 0 : µ = µ 0 X µ < µ 0 (3 ) 17 H 0 : µ = ν 0 X ( ) 61
3.4.2 2 µ x, µ y (1) 2 (2) 2 (3) 2 (4) 2 (1) 2.2.4 2 N(µ X, σ 2 X ), N(µ Y, σ 2 Y ) 2 m, n X, Ȳ N(µ X, σx 2 /m), N(µ Y, σy 2 /n) 2.2.5 X Ȳ N (µ X µ Y, µ X = µ Y X Ȳ N (0, σx 2 m + σ2 Y n σx 2 ) m + σ2 Y n ) 62
(2) t 2 N(µ X, σ 2 ), N(µ Y, σ 2 ) 2 m, n X, Ȳ U 2 X, U 2 Y T = X Ȳ (µ X µ Y ) V ( 2 1 m + ), V 2 = (m 1)U X 2 + (n 1)U Y 2 1 m + n 2 n T m + n 2 t µ X = µ Y T = X Ȳ V ( 2 1 m + ), V 2 = (m 1)U X 2 + (n 1)U Y 2 1 m + n 2 n T m + n 2 t V t (3) Welch ν t x ȳ t = ˆσ 2 x n x + ˆσ2 y n y ˆσ x, 2 ˆσ y 2 x, y ν ν 1 ν = c2 (1 c)2 + n x 1 n y 1, c = ˆσ 2 x n x ˆσ x 2 n x + ˆσ2 y n y t t > t ν (α) 63
H 0 : µ x = µ y H 0 (4) n 1 t t = x ȳ Ud 2 n U 2 d x i y i 3.4.3 ( ) p = 0.01 p < 0.01 p p = 0.01 Z = X 1 + + X n X i p p(1 p) = n p(1 p) ( 2 + (1 p)(0 p) 2 ) 2.2.5 0.01(1 0.01) N 0.01, ( n p = 0.01 ) P (Z < z(α)) = α z(α)( 100α ) Z < z(α) Z α α p < 0.01 Z z(α) α ( p = p 1, (p 1 > p 0 = 0.01) N p 1, p ) 1(1 p 1 ) p 1 0.5 p = p 0 = 0.01 n 64
100α p = p 0 = 0.01 p = p 1 1. H 0 : p = p 0 p = p 1 p 0 2. p 0 0.5 3. p 0 0.5 p 0 p 0 2 (p < p 0?) p 0 100 2 5 p 0 = 0.05 p p 0 p = p 0 p = p 0 p 0 k n k B(n, p) p = p 0 B(n, p 0 ) p 0 n B(n, p 0 ) P (k < K(α)) = α (1) K(α) 2 α K(α) (1) K(α) 65
K ( ) K n P (k K) = p k 0(1 p 0 ) n k (2) k k=0 P (2) p = p 0 (K p = p 0 ) p < p 0 p > p 0 K/n < p 0 p > p 0 p p 0 (2) (3) p = p 0 p > p 0 (3) d d p pk (1 p) n k = kp k 1 (1 p) n k (n k)p k (1 p) n k 1 = p k 1 (1 p) n k 1 (k pn) p p k (1 p) n k k/n < p (2) k k K K/n < p (2) p k (1 p) n k p (3) P ( ) n P (k K) = p k 0(1 p 0 ) n k (4) k k=k K/n < p 0 p < p 0 p p 0 (4) (5) P P 2 2 66
F 2 F 2 B(n, p) P B(n,p) (m, n) F P F m n K P B(n,p) (k K) = P F 2(K+1) 2(n K) ( (n K)p ) (K + 1)(1 p) < F K ( ) P F P p = p 0 α P α p = p 0 p < p 0 p F P F α (m, n) F 100α F m n (α) Z = K n H 0 : p = p 0 (n K)p ( 0 (K + 1)(1 p 0 ) F 2(K+1) α ) 2(n K) 2 K(1 p 0 ) ( F 2(n K+1) α ) 2K (n K + 1)p 0 2 H K(1 p 0 ) F 2(n K+1) 2K (α) (n K + 1)p 0 H H 3.4.3. (n K)p 0 (K + 1)(1 p 0 ) F 2(K+1) 2(n K) (α) 1. 5 67
5 150 2 5 5 2 P P B(0.05,150) (k 2) = 0.0181 2 p = 0.05 p < 0.05 2. 100 0.5 1 3.4.4 18 67 85 45 65 110 63 132 195 68
x a x a b x n a b + x n a b n b n n a b X ( ) n b n b n b P (X = x) X = x n b a x n a b k X = x ( ) ( ) a n a P (X = k) = k b k ( ) n b X x P x P (X = k) k=0 69
( n, a, b) 2 a b n 1 = a + b c d n 2 = c + d a + c b + d n = a + b + c + d q 1 = a/n 1 q 2 = c/n 2 q = (a + c)/n q 1, q 2, q p 1, p 2, p p 1 = p 2 (= p) X 1 1 0 X 2 1 0 E(X 1 ) = E(X 2 ) = p V (X 1 ) = V (X 2 ) = p(1 p) q 1, q 2 X 1, X 2 2.2.5 p(1 p) p(1 p) p n 1 n 2 q 1 q 2 0 ( 1 v = + 1 n 1 n 2 ) p(1 p) p q 2 t q 2 70
2 u u (q 1 q 2 ) 2 v (q 1 q 2 ) 2 p = q v (q 1 q 2 ) 2 v = n(ad bc) 2 (a + b)(c + d)(a + c)(b + d) 3 a b c n 1 = a + b + c d e f n 2 = d + e + f a + d b + e c + f n = a + b + c + d + e + f 3 3 m n 2 Fisher 2 R R 18 67 85 35 65 100 53 132 185 71
R chisq.test Fisher R fisher.test chisq.test 2x2 Yates correct=f with Yates continuity correction Yates 2x2 P Yates > chisq.test(matrix(c(18,67,35,65),ncol=2),correct=f) Pearson s Chi-squared test data: matrix(c(18, 67, 35, 65), ncol = 2) X-squared = 4.2952, df = 1, p-value = 0.03822 > chisq.test(matrix(c(18,67,35,65),ncol=2)) Pearson s Chi-squared test with Yates continuity correction data: matrix(c(18, 67, 35, 65), ncol = 2) X-squared = 3.6455, df = 1, p-value = 0.05622 > fisher.test(matrix(c(18,67,35,65),ncol=2),alternative= t ) Fisher s Exact Test for Count Data data: matrix(c(18, 67, 35, 65), ncol = 2) p-value = 0.04994 alternative hypothesis: true odds ratio is not equal to 1 95 percent confidence interval: 0.2412028 1.0134738 sample estimates: 72
odds ratio 0.5007994 > fisher.test(matrix(c(18,67,35,65),ncol=2),alternative= less ) Fisher s Exact Test for Count Data data: matrix(c(18, 67, 35, 65), ncol = 2) p-value = 0.0275 alternative hypothesis: true odds ratio is less than 1 95 percent confidence interval: 0.0000000 0.9144743 sample estimates: odds ratio 0.5007994 3.4.5 40 y 11 y 12 y 13 y 1 = y 11 + y 12 + y 13 1 39 y 21 y 22 y 23 y 2 = y 21 + y 22 + y 23 y 31 y 32 y 33 y 3 = y 31 + y 32 + y 33 y 1 y 2 y 3 y = y 1 + y 2 + y 3 73
2 (X) X 2 3 1 t F p R fisher.test 74
2 a b a = b = 3 χ 2 χ 2 = ij (y ij yˆ ij ) 2 (1) yˆ ij ν = (a 1)(b 1) yˆ ij y ij y 11 : y 12 : y 13 = y 21 : y 22 : y 23 = y 31 : y 32 : y 33 y 1 : y 2 : y 3 y 21 = y 2 y 1 y 11 y 11 + y 21 + y 31 = y 1 y 11 + y 2 y 1 y 11 + y 3 y 1 y 11 = y 1 y 11 yˆ 11 yˆ 11 = y 1 y 1 y 1 + y 2 + y 3 y i y j yˆ ij = (2) y 1 + y 2 + y 3 ( yˆ ij ) (a 1)(b 1) α χ 2 y ij α 75
α (1) χ 2 (3) y ij = f ij, yˆ ij = e ij 3.4.1. m n {f ij } 1 i m, 1 j n f i = j f ij, f j = i f ij, f = i,j f ij (f ij e ij ) 2 = e ij i,j ( ) i,j f ij e ij = f i f j f. i,j f 2 ij f i f j 1 (3) (f ij e ij ) 2 e ij = i,j ( ) 2 f ij fi f j f f i f j f = (f ijf f i f j ) 2 f f i f j = f 2 ij f 2 2f ij f f i f j + f 2 i f 2 j f f i f j = f f 2 ij f i f j 2f ij + f i f j f f i f j = ij i e ij f i j i,j f j = f 2 (f ij e ij ) 2 fij 2 = f f f i f j = f i,j f 2 ij 1 f i f j 76
40 20 49 10 79 1 39 6 20 41 67 0 12 34 46 26 81 85 192 (2) 40 10.7 33.3 35.0 79 1 39 9.1 28.3 29.7 67 6.2 19.4 20.4 46 26 81 85 192 (1) χ 2 59.3 P 4.132 10 12 = (3 1)(3 1) = 4 0.01 13.3 1 R R > chisq.test(matrix(c(20,49,10,6,20,41,0,12,34),ncol=3,byrow=t)) Pearson s Chi-squared test data: matrix(c(20, 49, 10, 6, 20, 41, 0, 12, 34), ncol = 3, byrow = T) X-squared = 59.2687, df = 4, p-value = 4.132e-12 > fisher.test(matrix(c(20,49,10,6,20,41,0,12,34),ncol=3,byrow=t)) Fisher s Exact Test for Count Data data: matrix(c(20, 49, 10, 6, 20, 41, 0, 12, 34), ncol = 3, byrow = T) 77
p-value = 1.071e-13 alternative hypothesis: two.sided 3.5 1. Excel Excel2007 (a) ( ) 165,162,171,... (b) X average( ) n counta( ) (c) µ µ 0 (d) Excel (a) [ -d, +d] d -norminv(α/2, 0, sqrt( / )) (b) P(X<norminv(p, m, v))=p p<0.5 m=0 norminv (c) z(α/2) norminv α/2 (d) norminv( /2, 0, sqrt( / )) (e) P normdist(-abs( - ),0,sqrt( / ), 1)*2 (f) normdist(*, *, 1) abs (g) P normdist 2 (h) norminv(, 0, sqrt( / )) (i) P normdist(-abs( - ), 0, sqrt( / ), 1) 78
t (a) [ -t*sqrt( / ), +t*sqrt( / )] t tinv(α, -1) (b) var( ) (c) P( X >tinv(p, n))=p tinv α t(α) (d) tinv(, -1) (e) P tdist(abs( - ), 0, 2) (f) tdist 3 2 P (g) tdist abs (h) tinv(2*, -1) (i) tinv t(α) 2 (j) P tdist(abs( - ),0,1) (k) tdist 3 1 P (l) tdist abs 2. R R R R Enter help( 79
> data <- c(3,5,4,6,5,7,3,8,4,5) > n <- length(data) > s <- sd(data) > x_bar <- mean(data) sd(data) data qnorm(p) P (X z) = p z pnorm(z) P (X z) > z <- qnorm(0.975,mean=0,sd=s/sqrt(n)) > c(x_bar-z, x_bar+z) [1] 3.987879 6.012121 > 2*(1-pnorm(abs(6-x_bar),mean=0,sd=s/sqrt(n))) [1] 0.05280751 5% t 6 P t t qt(p) t P (X t) = p t pt(x) t P (X t) > t <- qt(0.975,df=n-1)*s/sqrt(n) > c(x_bar-t, x_bar+t) [1] 3.831827 6.168173 > 2*(1-pt(abs(6-x_bar)/(s/sqrt(n)),n-1)) [1] 0.08478521 5% 6 80
R > t.test(data,mu=6) One Sample t-test data: data t = -1.9365, df = 9, p-value = 0.08479 alternative hypothesis: true mean is not equal to 6 95 percent confidence interval: 3.831827 6.168173 sample estimates: mean of x 5 > t.test(data,mu=6,alternative="less") One Sample t-test data: data t = -1.9365, df = 9, p-value = 0.04239 alternative hypothesis: true mean is less than 6 95 percent confidence interval: -Inf 5.946615 sample estimates: mean of x 5 2 R 81
> data2 <- c(4,4,5,6,5,7,6,6,9,6) > t.test(data,data2,var.equal=f) Welch Two Sample t-test data: data and data2 t = -1.1494, df = 17.819, p-value = 0.2656 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -2.2633529 0.6633529 sample estimates: mean of x mean of y 5.0 5.8 3. R 2 100 55 2 pbinom(k,n,p) P B(n, p) X P (X k) > 1-pbinom(54,100,0.5) [1] 0.1841008 P B(100, 0.5) X P (55 X) = 1 P (X 54) binom.test > binom.test(c(55,45),p=0.5,alternative="greater") Exact binomial test data: c(55, 45) number of successes = 55, number of trials = 100 p-value = 0.1841 82
alternative hypothesis: true probability of success is greater than 0.5 95 percent confidence interval: 0.4628896 1.0000000 sample estimates: probability of success 0.55 4. R data3.txt Excel 20 49 10 6 20 41 0 12 34 R data3.txt > data3<-read.table("data3.txt") > data3 V1 V2 V3 1 20 49 10 2 6 20 41 3 0 12 34 > chisq.test(data3) Pearson s Chi-squared test data: data3 X-squared = 59.2687, df = 4, p-value = 4.132e-12 R R 83
R 1. R Windows R R 2. R > 3. 3,4,5 d > d <- c(3,4,5) <- < - = <- Enter 4. d d > d Enter [1] 3 4 5 5. R Enter 84
4 4.1 1 50 C 55 C 60 C 65 C 77.4 78.3 79.2 78.9 78.2 78.2 79.3 78.8 78.1 78.4 79.1 78.1 77.8 77.3 78.2 78.1 77.9 79.1 79.3 78.9 ( ) ( ) ( ) F A( ) a ( ) i r i i j y ij ( r i 5 ) y ij = µ i + ε ij 1 85
µ i i ε ij N(0, σ 2 )( ) H 0 : µ 1 = µ 2 = = µ a y i = j y ij, y = i j y ij ȳ i = y i /r i, ȳ = y /n S e = (y ij ȳ i ) 2 i j S e /σ 2 ν e = n a S T = (y ij ȳ ) 2 i j S e S A = S T S e S A /σ 2 (S e ) H 0 ν A = a 1 ( S A ) H 0 F = S A/ν A S e /ν e, ν A = a 1, ν e = n a ν A, ν e F µ i F F α f νa,ν e (α) F > f νa,ν e (α) H 0 F (Analysis of Variance, ANOVA) 86
4.2 1. Excel Excel (a) Excel (b) (c) ( ) (d) F F F > F 2. R R 2 1 >score=c(56,48,72,60,55,60,62,76,84,78,53,62,44,90,57, 77,72,83,81,91,83) > group=factor(rep(c("a", "B", "C", "D"), c(5, 4, 6, 6))) > anova(aov(score ~ group)) Analysis of Variance Table Response: score Df Sum Sq Mean Sq F value Pr(>F) group 3 1629.2 543.06 3.9141 0.02708 * Residuals 17 2358.6 138.74 87
--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05. 0.1 1 score (group) group A,B,C,D 5,4,6,6 A 56,48,72,60,55 B 60,62,76,84 residual group - Df Signif. codes *** p 0 0.001 ** 0.001 0.01 * 0.01 0.05. 0.05 0.1 0.1 * 0.01 0.05 > score=c(5,6,7,6,7,8,7,8,9) > group=factor(rep(c("a","b","c"),c(3,3,3))) > anova(aov(score ~ group)) Analysis of Variance Table Response: score Df Sum Sq Mean Sq F value Pr(>F) group 2 6 3 3 0.125 Residuals 6 6 1 4.3 x 1,..., x p, y (x 11,..., x p1, y 1 ),..., (x 1n,..., x pn, y n ) y x 1,..., x p y = b 0 + b 1 x 1 + + b p x p x 1,..., x p y f(x 1,..., x p ) = b 0 + b 1 x 1 + b p x p 1. 88
2. b 0,..., b p S = (y i f(x 1i,..., x pi )) 2 S 3. b 0,..., b p S b i = 0 (i = 0,..., p) 0 (b 0,..., b p ) S y = b 0 + b 1 x 1 + + b p x p S 4.3.1 S b 0,..., b p (b 0,..., b p ) S b k S b k = 2(y i b 0 b 1 x 1i b p x pi )( x ki ) = 0, k = 0,..., p x 0i = 1 ( n ) ( n ) ( n ) x ki b 0 + x ki x 1i b 1 + + x ki x pi b p = x ki y i, k = 0,..., p b 0,..., b p b k p = 1, p = 2 p = 1 nb 0 + ( ) x i b 0 + 89 ( ) x i b 1 = n y i ( ) b 1 = n y i x i 2 x i
ns 2 x = x 2 i n x 2, ns xy = x i y i n xȳ x 2 b 0 = i yi x i xi y i n x 2 i ( x i ) 2 = nȳ x 2 i n x x i y i + n 2 x 2 ȳ n 2 x 2 ȳ n 2 s 2 x = r n2 ȳs 2 x n 2 xs xy n 2 s 2 x = ȳ x s xy s 2 x b 1 = = x 2 i yi x i xi y i n x 2 i ( x i ) 2 (xi x)(y i ȳ) (xi x) 2 p = 2 nb 0 + ( ) x 1i b 0 + ( ) x 2i b 0 + = r s y s x ( ) x 1i b 1 + ( ) b 1 + ( x 2 1i ) x 2i x 1i b 1 + ( ) x 2i b 2 = n y i ( ) x 1i x 2i b 2 = n x 1i y i ( ) b 2 = n x 2i y i 2 x 2i y = f(x 1,..., x p ) = b 0 + b 1 x 1 + + b p x p b k 4.3.2 X 1 x 11 x p1 y 1 b 0 1 x 12 x p2 X =..., y = y 2., b = b 1. 1 x 1n x pn y n b p 90
n x1i x2i 2 x1i x1i x1i x 2i X X = 2 x2i x2i x 1i x2i..... xpi xpi x 1i xpi x1i x pi. xpi 2 X Xb = X y X V = {Xb : b R p+1 } X V u 0,..., u r V (V X X - ) y V y u v V u V u u = r i=0 (y, u i )u i v + w 2 = (v + w, v + w) = v 2 + 2(v, w) + w 2 v V v u y u y v 2 = v u 2 + y u 2 v = Xb, u = Xˆb y v 2 = S(b), y u 2 = S(ˆb) ˆb S x j, j = 0,..., p X (y Xˆb, v) = 0, v V (y Xˆb, x j ), j X (y Xˆb) = 0 Xˆb y V ˆb S 91
X X (X X) X X(X X) X X = X X X X (X X) X y ( ) S ˆb = (ˆb0,, ˆb ) p b = (b 0,, b p ) S(b) = (y Xb) (y Xb) { } { } = y Xˆb + X(ˆb b) y Xˆb + X(ˆb b) (4.1) = (y Xˆb) (y Xˆb) + (Xb Xˆb) (Xb Xˆb) (4.2) S(ˆb) (Xˆb Xb) (y Xˆb) (y Xˆb) (Xˆb Xb) X (y Xˆb) = 0 4.3.3 ( ) 1. y x 1,..., x p y = β 0 + β 1 x 1 + + β p x p x k y 2. y x 1,..., x p (x 1i,..., x pi ) 2 y i = β 0 + β 1 x 1i + + β p x pi + ε i ε i i 1 ( ) 2 92
3. ε i N(0, σ 2 ) (σ ) y i ε i y = b 0 + b 1 x 1 + + b p x p b k b k β k E(b k ) = β k b k β k b 0,..., b p l 0 b 0 + + l p b p l 0,..., l p 1, x,..., x p 1, x 1,..., x p ( ) S l 0,..., l p 1, x,..., x p 4.3.1 (Gauss-Markov ). l 1 b 1 + + l p b p ȳ l 1ˆb1 + + l pˆbp (1) - F 0 t 0 0 H 0 : β 1 = = β p = 0 93
( ) S e = (y i ŷ i ) 2, S R = S yy = (y i ȳ) 2 S e = S yy S R (ŷ i ŷ) 2 ( H 0 ) S e /σ 2 N(0, 1) ν e 2 ( p.56) ν e = n rank(x)(x n (p + 1) ) ν e = n p 1 χ 2 S e /σ 2 n p 1 χ 2 S R /σ 2 H 0 N(0, 1) p 2 ( p.56 ) χ 2 S R /σ 2 p χ 2 F H 0 F = S Rp S e n p 1 (p, n p 1) F F 94
S R ( ) Syy /(n p 1) p F = 1/ S R /p n p 1 F α f νa,ν e (α) F > f νa,ν e (α) H 0 V e = S e /(n p 1) ( H 0 )σ 2 V R = S R /p σ 2 F (2) - (y i b 0 b 1 x 1 b p x p ) 2 R 2 = 1 (y i ȳ) 2 R y y i b 0 + b 1 x 1i +... + b p x pi 4.3.2. y b 0 + b 1 l 1 + + b p l p ( p.16 ) F F 95
4.3.3. t k = b k β k, a kk V e k = 1,..., p b 0 β 0 t 0 = ( ) 1 n + p p x j x k a jk V e j=1 k=1 n p 1 t V e V e = (y i f(x 1,..., x p )) 2 n p 1 a jk ( n ) A = (a jk ) = (x ji x j )(x ki x k ) (j, k) ( ) ( p.60 ) (β k = 0) P (t k > t n p 1 (α)) = P (t k < t n p 1 (α)) = α 2 t n p 1 (α) 1 α β k [ b k a kk V e, b k + a kk V e ], k = 1,..., p b 0 1 p n + j=1 k=1 p x j x k a jk V e, b 0 + 1 p n + j=1 k=1 p x j x k a jk V e, k = 0 96
H 0 : β k = 0 x k x k b k k = 1,..., p k = 0 t k = b k a kk V e, k = 1,..., p b 0 t 0 = ( ) 1 n + p p x j x k a jk V e j=1 k=1 b k 0 t k > t n p 1 (α) α H 0 4.4 1. data mulreg.xls 2. 3. OK 4. Y ( ) X ( ) ( ) OK 5. (F ) t 6. 97
R 2 R 2 = 1 (y i b 0 b 1 x 1 b p x p ) 2 (y i ȳ) 2 = 1 S e S yy S e, S yy R 2 R 2 R 2 R 2 = 1 S e/(n p 1) S yy /(n 1) 4.5 4.5.1 p x 1,, x p x 1 x 11 x 1i x 1n.. x k x k1 x ki x kn.... x p x p1 x pi x pn p = 2 x 1 x 2.. 98
x 1,..., x p x 1,..., x p ( ) (a 1,..., a p ), a 2 1 +... + a 2 p = 1 (x 1,..., x p ) z z = a 1 x 1 +... + a p x p z z (x 1i,..., x pi ) z z i (a 1,..., a p ) V (z) = 1 (z i z) 2, z = 1 n n V (z) z (a 1,..., a p ) V (z) x 1,..., x p, y, a 1,..., a p V (z) = 1 n = 1 n = 1 n (a 1x 1i + + a p x pi ) 1 n a 1 x 1i 1 n j=1 x 1j z i (a 1 x 1j + + a p x pj ) j=1 + + a p x pi 1 n {a 1 (x 1i x 1 ) + + a p (x pi x p )} 2 j=1 x pj 2 2 x i = 1 n n j=1 x ij σ ij = 1 (x ik x i )(x jk x j ) n k=1 (σ ij x i, x j ) V (z) = σ 11 a 2 1 + + σ pp a 2 p + 2 i<j σ ij a i a j 99
a 2 1 + + a 2 p = 1 a 1,..., a p σ 11 a 2 1 + + σ pp a 2 p + 2 i<j σ ij a i a j a 1,..., a p σ ii a i + j i σ ij a j λa i = 0 i = 1,..., p a 1,..., a p, λ σ 11 σ 1p Σ =... σ p1 σ pp, α = (Σ ) Σα = λα (1) p p 1 (positive semi-definite) Σ (λ 1, α 1 ),..., (λ p, α p ), α 1,..., α p 1 a 1. a p λ 1 λ 2... λ p (λ, α) Σ V (z) = α Σα V (z) = α Σα = α λα = λα α = λ 100
λ 1 α 1 V (z) z α 1 = (a 11,..., a 1p ) z 1 = α 1 (x 1,..., x p ) = a 11 x 1 + + a 1p x p α 1 λ 1 λ 2 λ 1 α 2 = (a 21,..., a 2p ) z 2 = α 2 (x 1,..., x p ) = a 21 x 1 + + a 2p x p α 2 α 1 α = 1 α α 1 = 0 V (z) Lagrange (1) α = 1 α 1 V (z) α (1) α = α 2 R.A.Johnson, D.W.Wichern, Applied Multivariate Analysis, sixth edition, Chpter 8.1 p 4.5.2 ( ) p σ ii p σ ii Σ Σ 101
k z k λ k (k ) C k = λ k p i σ = λ k p ii i λ i k k k Ci i k k z k x j z k a k ( ) z k λk a kj r kj = σjj 4.5.3 x 1,..., x p x i = x i x i σii x i x i 4.6 1. 102
2. Excel ( ) OK 3. 4. biplot 103