Scarica FORMULARIO STATISTICA e più Appunti in PDF di Statistica solo su Docsity!
FORMULARIO
PROBABILITA’
P (Ei|A) =
P (A|Ei) · P (Ei) ∑m j=1 P^ (A|Ej^ )^ ·^ P^ (Ej^ )
E (X) = μ =
x xP^ (x)^ μ^ =^ E^ (X) =^
−∞ xf^ (x)^ dx
σ 2 = V AR (X) = E [(X − μ)]
2
x (x^ −^ μ)
2 P (x) V AR (X) =
−∞ (xi − μ)
2 f (x) dx
X ∼ Ber (p) P (X = x) = px^ (1 − p)
1 −x E (X) = p V AR (X) = p (1 − p)
X ∼ Bin (n , p) P (X = x) =
n x
px^ (1 − p)
n−x E (X) = np V AR (X) = np (1 − p)
X ∼ Hg (N, s, n) P (X = x) =
s x
N − s n − x
N
n
E (X) =
ns
N
V AR (X) = np (1 − p)
N −n N − 1
X ∼ P o (λ) P (X = x) =
e −λ λ x
x!
E (X) = λ V AR (X) = λ
X ∼ N
μ , σ 2
f (x) =
2 πσ^2
e
− (^12) ( x−σμ ) 2 E (X) = μ V AR (X) = σ 2
Z =
X − μ
σ
X ∼ χ 2 (r) E^ (X) =^ r^ V AR^ (X) = 2r
X ∼ t(r) E (X) = 0 V AR (X) = r r− 2
X ∼ F(rX , rY )
VX /rX VY /rY =^
χ^2 (rX ) /rX χ^2 (rY )
/rY ∼^ F(rX ,rY )
P ∼ Bin (n , p) P ∼ N
p,
p (1 − p)
n
per n grande
X ∼ N
μ,
σ 2
n
(n − 1) S 2
σ^2
∼ χ 2 (n−1) Z^ =^
X − μ
σ/
n
∼ N (0, 1) T =
X − μ
S/
n
∼ t(n−1)
S^2 =
Xi − X
n − 1
X − Y
∼ N
μX − μY ,
σ^2 X
nX
σ^2 Y
nY
σ 2 X , σ
2 Y note^ Z^ =
X − Y
− (μX − μY ) √ σ^2 X nX +^
σ^2 Y nY
∼ N (0, 1)
σ 2 X =^ σ
2 Y incognite^ T^ =
X − Y
− (μX − μY ) √( (nX −1)s^2 X +(nY −1)s^2 Y nX +nY − 2
1 nX
1 nY
) ∼^ t(nX +nY −2)
INTERVALLI DI CONFIDENZA
X ∼ N
μ =?, σ^2
P
[
x − zα/ 2
σ √ n
≤ μ ≤ x + zα/ 2
σ √ n
]
= (1 − α)
X ∼ N
μ =?, σ 2 =?
P
[
x − tn− 1 ,α/ 2
s √ n
≤ μ ≤ x + tn− 1 ,α/ 2
s √ n
]
= (1 − α)
X ∼ N
μ, σ 2 =?
P
[
(n − 1) s^2
χ 2 (n−1),α/ 2
≤ σ 2 ≤
(n − 1) s^2
χ 2 (n−1),(1−α/2)
]
= (1 − α)
Pˆ ∼ Bin (n, p) P
[
p ˆ − zα/ 2
pˆ(1−pˆ) n ≤ p ≤ pˆ + zα/ 2
pˆ(1−pˆ) n
]
= (1 − α) n grande
Distribuzione congiunta
P
[
d^ ¯ − t n− 1 ,α/ 2
∑(d i−^ d¯)^2 n− 1 n ≤ μd ≤ d¯ + tn− 1 ,α/ 2
∑(d i−^ d¯)^2 n− 1 n
]
Normale di X e Y^ = (1^ −^ α)
X ∼ N
μX =?, σ 2 X
P
[
(x − y) − zα/ 2
σ X^2 nX +^
σ Y^2 nY ≤^ (μX^ −^ μY^ )^ ≤^ (x^ −^ y) +^ zα/^2
σ^2 X nX +^
σ^2 Y nY
]
= (1 − α)
Y ∼ N
μY =?, σ 2 Y
X ∼ N
μX =?, σ 2 X =?
P
[
(x − y) − t(nX +nY − 2 ,α/2)
s^2 p nX +^
s^2 p nY ≤^ (μX^ −^ μY^ )^ ≤^...
Y ∼ N
μY =?, σ 2 Y =?
· · · ≤ (x − y) + t(nX +nY − 2 ,α/2)
s^2 p nX +^
s^2 p nY
]
= (1 − α)
σ^2 X = σ Y^2 s^2 p =
(nX − 1) s^2 X + (nY − 1) s^2 Y
nX + nY − 2
X ∼ N
μX =?, σ^2 X =?
P
[
(x − y) − t(v,α/2)
s^2 X nX
s^2 Y nY ≤ (μX − μY ) ≤ (x − y) + t(v,α/2)
s^2 X nX
s^2 Y nY
]
= (1 − α)
Y ∼ N
μY =?, σ^2 Y =?
σ 2 X 6 =^ σ
2 Y v^ =
[(
s^2 X nX
s^2 Y nY
)] 2
s^2 X nX
/ (nX − 1) +
s^2 Y nY
(^2) / (nY − 1)
STATISTICA TEST CHI-QUADRATO
χ 2 =
∑^ k
i=
(Oi − Ei)
2
Ei
∼ χ 2 (k−1) χ
2
∑^ r
i=
∑^ c
j=
(Oij − Eij )
2
Eij
∼ χ 2 [(r−1)(c−1)]
Disuguaglianza di Cebycev: P {|X − μ| < εσ} ≥ 1 − 1 ε^2 P {|X − μ| < k} ≥ 1 − σ
2 k^2
