Tests d'hypothèses, Intervalle de con ance

10 000
7,5
95 10 000
95
10 000 [p−yαqp(1−p)
n;fe +
yαqp(1−p)
n]
150
p= 0.75 n n > 150
X n
X
95
n P [X > 150] ≤0.05
300
p= 0.5p= 0.8
X
n, p E(X)=0.75n X = 0.25 ·0.75n n > 150
0,75n
σ=√0.25 ·0.75n P [X > 150] ≤0.05 P[X≤150] ≥0.95
P[X−0.75n
√0.25·0.75n≤150−0.75n √0.25·0.75n]≥0.95 F(1.645) =
0.95 150.5−0.75n
√0.25·0.75n≥1.645
0≤n≤187.
187
5
N= 150 p= 0.5n150.5−0.5n
√0.5.0.5n≥1.645
n≤272
N= 300 p= 0.75 n300.5−0.75n
√0.25.0.75n≥1.645
n≤381
N= 300 p= 0.5n300.5−0.5n
√0.5.0.5n≥1.645
n≤561

X
X
X n = 30 p= 0.2
n
meσe
m
σ
X
Y
m68
σ σeq300
299 = 7q300 299 '7.0117
m Iα= [67.2; 68.8]
69
70
26.2 300
XiYi
XiYi

Z=P300
i=1(Xi+Yi) 600
E(Z) = 300 ·(70 + 15) = 25 500
Z Z = 300 ·(Xi+Yi)Z
m= 25 500
σ=p300 ·(82+ 52) = 163.4Z0=Z−m
σ
Z > 26 200 Z0>4.284
t= 4 F(t)=0.999 968 = P[Z0≤4]
0.000 04
4
240
1
20
4
n= 240 p=1
20
n≥30 np ≤15,
λ np = 12
200
50
12
Iα= [11; 13]

α= 0.05
H0
p0= 0.1
p
f
Xii≤50,
1 0
H0n= 50 p0= 0.1
H0p=p0= 0.1
X=Pi=50
i=1 Xi
n
10
50
p0
qp0(1−p0)
50
I'[0.017; 0.183]
0.2
95 H00.05
H0

C
F
F C P (F∩C) = P(F)·
P(C)P(C∩F), P (C∩
F), P (C∩F)H0
P(F) = 600
1000 P(C) = 500 1000 P(F)·P(C) = 3 10 300
H0
s=Pi=4
i=1
(Oi−Ti)2
Tis= 34.73
α= 0,001
P[χ2≥10.83] = 0.001 χ2
34.73
H0
1000
χ2
10.83 χ2
34.73