(normal distribution) 30 2
Skewed graph 1
2 (variance) s 2 = 1/(n-1) (xi x) 2 x = mean, s = variance
(variance) (standard deviation) SD = SQR (var) or
8 8
0.3 0.2 0.1 0.0 0 1 2 3 4 5 6 7 8 8 0 1 8 (probability distribution
8 20
0.20 0.15 0.10 0.05 0.00 0 2 4 6 8 10 12 14 16 18 20 4 20 0 1 20 8
(normal distribution/gaussian distribution/bellshaped distribution) m standard deviation (s) m 0 s 1 2SD, 1SD 2.3%, 15.9% p p 0.05
SD:15.9% 2SD 2.3% - 2SD -1SD 1SD 2SD z score 1SD 2SD 100 % 15.9%, 2.3%
(z score) 1.645 1.96 0.05
5% 2.5% -1.96-1.64 1.64 1.96 z score 6 100 % 5%, 2.5% (= z score) 1.645, 1.96
5.0 SD 0.5 3.0 SD 1 SD Z = (X 2)/0.5 Z = (5 2)/0.5 = 6 4 100cm 10cm 80cm SD (80 100) / 10 = 2SD 18 74 129mmHg 19.8 2.5 2.5 z=1.96 1.96 (X 129)/19.8 X=167.8 mmhg 2.5 168mmHg 97.5 1.96=(X + 129)/19.8 X = 90.2 mmhg 1.96SD 2.5 150mmHg Z = 150 129 / 19.8, z=1.06, 14.5% 14.5 150mmHg
SD0.5 SD1.0 0 2 mean 7 100 % 5%, 2.5% (= z score) 1.645, 1.96
2.5% 2.5% 90.2 129 167.8 (mm Hg) z = (X µ) σ / n µ = X = σ = n = 8 100 % 5%, 2.5% (= z score) 1.645, 1.96
inference X m Maximum likelihood estimator 2 20 70 60 2 x1 x2 sampling distribution sampling confidence interval (CI)
95% CI 193 229 9 100 95 (confidence interval: CI) 95
A 10 3 30% 95% 0.3 sqr[0.3(1-0.3)]/10] 0 0 0.6 10 A 6 10 3 95% 29.7 30.3% 30% A 95% 29 7% 30 3% A
211 mg/dl 25 220 mg/dl 220 25 25 5 5 5 2.5% 1.645 1.96 1.96SD 25 Ho 2 H0 accept 25 p < 0.05
(= 46) z =(X - µ 0 )/ s / n z =(220-211) / 46 / 25 = 0.98 1.645 1.96 25 1.96 =( - 211) / 46 / 25 x 229 mg/dl
µ 0 25 µ 1 211 220 (mg/dl) Null hypothesis H 0 : µ 0 = µ 1 Alternative hypothesis H 0 : µ 0 µ 1 10 211 25 220
Binomial Distribution Yes/No 0.29 0.71 0.71 2 2 1 1 0.29 0.71 0.71 0.29 11 3 12 13
Yes/No = 29%, = 71% : (0.29) 2 : (0.71) 2 1 : (0.29)(0.71)x2 1.0 11
3 0.71 3 x 3 C 3 = 0.3579 2 1 0.71 2 x 0.29 x 3 C 2 = 0.4386 1 2 0.71 x 0.29 2 x 3 C 1 = 0.1791 3 0.29 3 x 3 C 0 = 0.244 1 7 3 7C 3 = 7 x 6 x 5 3 x 2 x 1 12
P(X=x) = n C x p x (1-p) n-x Mean = np = 0.29 x 10 = 2.9 SD = np(1-p) = 2.059 = 1.4 13
10 (p=0.000) (p=0.0326) 10 6 0.0404 0.05 10 3 1 10 0 10 14 (variance) np(1-p)
10 (p=0.000) (p=0.0326) 10 6 0.0404 0.05 10 3 1 10 0 10 14 (variance) np(1-p)
Binomial Distribution (skewed) 14
P 0.5 SD 0 1 SD 15
p = 0.5 p = 0.29 p = 0.71 15
1 20 3 5 20 3 20CK (0.05)k(0.95)20-k, K = 0, 1, 2,.... 20 3 0, 1, 2, 20C0 (0.05)0(0.95)20 = 0.3585 20C1 (0.05)1(0.95)19 = 0.3774 20C2 (0.05)2(0.95)18 = 0.1887 1 (0.3585 + 0.3774 + 0.1887) = 0.0754 20 3 7.5 cut off 5 3 3
X 0.00001 2,500,000 36 X 0.05 Pr(k >= 36) 20 1 14 1 20 14 1 0.04 0.05 20 3
16
8 4 active, 5 inactive 0.2 active 0.0, 0.1, 0.2,... 0.9, 1.0 accept P 0.2 active (0.99)
Operating Characteristic Curve (OC) N=8, p = 0.2 Pregnant probability of accept cumulative 0. 0.1678 0.1678 1. 0.3355 0.5033 2. 0.2936 0.7969 3. 0.1468 0.9437 4. 0.0459 0.9896 6. 0.0092 0.9988 7. 0.0011 0.9999 8. 0.0001 1.0000 9. 0.0000 17
18 19
Operating Characteristic Curve (OC) 0.0, 0.1, 0.2,... 0.9, 1.0 accept P=0 P=0.1 P=0.2 P=0.3 P=0.4 P=0. 5 P=0. 6 P=0. 7 P=0. 8 P=0. 9 P=1 0 1 0.43 0.17 0.06 0.02 0.00 0.00 0.00 0.00 0.00 0.00 1 0 0.81 0.50 0.26 0.11 0.04 0.01 0.00 0.00 0.00 0.00 2 0 0.96 0.80 0.55 0.32 0.14 0.05 0.01 0.00 0.00 0.00 3 0 0.99 0.94 0.81 0.59 0.36 0.17 0.06 0.01 0.00 0.00 4 0 1.00 0.99 0.94 0.83 0.64 0.41 0.19 0.06 0.01 0.00 5 0 1.00 1.00 0.99 0.95 0.86 0.68 0.45 0.20 0.04 0.00 6 0 1.00 1.00 1.00 0.99 0.96 0.89 0.74 0.50 0.19 0.00 7 0 1 1 1 1 1.00 0.98 0.94 0.83 0.57 0.00 8 0 1 1 1 1 1 1 1 1 1 1 18
8 4 accept accept 0 0.1 8 OK 0.2 0.3 0.4 6 accept 0.2 8 3 accept 0.2 94 accept 0.5 36 accept 19 Operating Characteristic Curve (OC) OC (two stage screening)
Operating Characteristic Curve (OC) 1 accept 0 0 true 1 Cf. two stage screening 19
Poisson distribution 0.00024 binomial situation binomial distribution Poisson person-time Poisson distribution 0.5 0 20 variance = np(1-p) Poisson p 0 1-p variance = np = mean 21
p 0, 1 p = 1, mean = variance = np Poisson Binomial 20
Poisson distribution 2 independence assumption B A Poisson Stationary assumption Poisson 1 1 Poisson Hazard model Poisson
Poisson Distribution P X x e λ λ x /x! 0 < x < infinity e=2.7182 p 0, 1 p 1 mean = variance = np 21
0.00024 1 4 l = np = 10,000 x 0.00024 = 2.4 P(X=4) = e-2.4 (2.4)4 / 4! = 0.1254 12.4 3 l = np = 3 P(X=x) = (x 3) / 3 > 1.645 (p=0.05) X = 6 6