Statistiques Inférentielles

X X
[a, b]
α X
α=P(X6∈ [a, b]).
P(X∈[a, b]) = 1 −α.
1−α
X
IF1
IF2
X IF3
IF4
XB(n, p)α]0,1[
n= 100 p= 0,3α= 0,05
IF1[22,39]
IF2np
n p [21,39]
IF3X[20,39]
X X
IF4
B N p
•IF1IF2
IF1IF2
•IF3X= 0
a P (X≤a)=0,025
b X =n
P(X≥b)=0,025 X= 0
b P (X≤b) = 0,975.
•n, p, α
a b IF3

IF X

IF4
[np −√n;np +√n]
[np −uαpnp(1 −p); np +uαpnp(1 −p)]
uαP(−uα< Z < uα)=1−α Z N(0,1)
n= 100 p= 0.3α= 0.05 uα= 1.96
IF4= [30 −10 ; 30 + 10] [21,018 ; 38,982]
p n
p
n
n
XnB(n, p)
P(Xn≤a)≈P(Zn≤a)
ZnN(np, pnp(1 −p))
•a P (Zn≤a) = 0.025 ZnN(np, pnp(1 −p))
•b P (Zn≤b) = 0.975 ZnN(np, pnp(1 −p))
N(np, pnp(1 −p))
0,025 0,975 a b np
(n∗p∗(1 −p))

N(0,1) ZnN(np, pnp(1 −p)) Z=Zn−np
√np(1−p)N(0,1)
P(Zn≤a) = P(Zn−np ≤a−np) = PZn−np
√np(1−p)≤a−np √np(1−p)=P(Z≤a0)a0=a−np √np(1−p)
a0P(Z < a0) = 0.025
ΦN(0,1) Φ−1(0.025) a0=−1.96
−uα−zα/2α= 0.05
N(0,1)
a−np
√np(1−p)=−1.96
a=np −1.96pnp(1 −p).
b P (Zn≤b) = P(Zn−np
√np(1−p)≤b−np √np(1−p)) = P(Z≤b0)
0,975 Φ−1(0.975) = 1.96 P(Z < 1.96) = 0.975
b=np + 1.96pnp(1 −p).
Xn
I= [np −1.96pnp(1 −p); np + 1.96pnp(1 −p)].
ppp(1 −p)≈1/2 1.96pp(1 −p)≤1
I= [np −√n;np +√n].
n
I=p−1
√n;p+1 √n.
XN(0,1) P(−2.576 <
X < 1.696) ≈0.95