Дронов С.В. Математические методы в психологии
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z
j
z
j
−
¯
X
S
Φ
z
j
−
¯
X
S
Φ(x)
p
j
np
j
n
j
(n
j
− np
j
)
2
(n
j
−np
j
)
2
np
j
χ
2
r −3
r
¯
X = 62, 45, S = 7, 48.
z
j
−∞ ∞
z
j
−
¯
X
S
−∞ ∞
Φ
z
j
−
¯
X
S
p
j
np
j
n
j
(n
j
− np
j
)
2
(n
j
−np
j
)
2
np
j
χ
2
= 6, 149
χ
2
χ
2
X n Y m
n + m
r
X n
1
n
r
n
Y m
1
m
r
m
n
1
+ m
1
n
r
+ m
r
n + m
χ
2
= (n + m)
r
X
i=1
n
2
i
n(n
i
+ m
i
)
+
r
X
i=1
m
2
i
m(n
i
+ m
i
)
− 1
!
r −1
X
Y
X
Y
χ
2
= 90(1, 117 − 1) = 10, 53
R(X, Y ) = 1 −
6
P
n
i=1
d
2
i
n
3
− n − T
X
− T
Y
.
X, Y d
i
i n
T
X
, T
Y
X, Y
T =
1
2
X
(t
3
j
− t
j
),
t
j
j T = 0
R
T
1
= T
2
= T
5
= 0
t
1
= 2 T
3
= (2
3
−2)/2 =
3 t
1
= 5 T
4
= (5
3
− 5)/2 = 60.
T = ((3
3
− 3) + (2
3
− 2))/2 = 15
R(1, 2) = 1 − 6×
×
(1−2)
2
+(3−1)
2
+(5−3)
2
+(7−4)
2
+(6−7)
2
+(4−6)
2
+(2−5)
2
7
3
−7−0−0
≈ 0, 41.
R(i, j) = R(j, i)
R(2, 5)
•
•
•
T
i
, i = 1, 2, 3
R
X
Y
Y = αX + β, X = kY + b.
α, β, k, b
S(α, β) =
n
X
i=1
(Y
i
− αX
i
− β)
2
→ min
α,β
T (k, b) =
n
X
i=1
(X
i
− kY
i
− b)
2
→ min
k,b
Y = ρ
S
Y
S
X
(X −
¯
X) +
¯
Y , X = ρ
S
X
S
Y
(Y −
¯
Y ) +
¯
X,
ρ S
X
, S
Y
X, Y
X Y
X r A
1
, ..., A
r
Y
s B
1
, ..., B
s
∆
i,j
A
i
B
j
X, Y ∆
i,j
X ∈ A
i
Y ∈ B
j
n
1
, ..., n
r
X
A
1
, ..., A
r
m
1
, ..., m
s
Y B
1
, ..., B
s
¯
Y
(i)
=
1
n
i
P
{k:X
k
∈A
i
}
Y
k
, i = 1, ..., r,
¯
X
(j)
=
1
m
j
P
{k:Y
k
∈B
j
}
X
k
, j = 1, ..., s.
(
˜
X
i
,
¯
Y
(i)
), i = 1, ..., r (
¯
X
(j)
,
˜
Y
j
), j = 1, ..., s
˜
X
i
A
i
˜
Y
j
B
j
Y X X Y
X
D
0
X X
X
D
D
0
X DX > D
0
X,
DX = D
0
X + D
A
X,
D
A
X
DX = D
0
X + D
A
X + D
B
X + D
A,B
X + ...
X
X
m n
X
i
j x
i,j
A
1
A
m
x
1,1
x
1,m
n x
n,1
x
n,m
¯x
1
¯x
m
N = nm
¯
¯
X =
1
m
m
X
j=1
¯x
j
=
1
nm
n
X
i=1
m
X
j=1
x
i,j
.
m
X
j=1
n
X
i=1
(x
i,j
−
¯
¯
X)
2
=
m
X
j=1
n
X
i=1
(x
i,j
− ¯x
j
)
2
+ n
m
X
j=1
(¯x
j
−
¯
¯
X)
2
.
Q
Q
0
n Q
A
Q X Q
0
Q
A
X
Q =
P
m
j=1
P
n
i=1
x
2
i,j
−
1
N
P
m
j=1
P
n
i=1
x
i,j
2
;
Q
0
=
P
m
j=1
P
n
i=1
x
2
i,j
−
1
n
P
m
j=1
(
P
n
i=1
x
i,j
)
2
;
Q
A
=
1
n
P
m
j=1
(
P
n
i=1
x
i,j
)
2
−
1
N
P
m
j=1
P
n
i=1
x
i,j
2
.
D
A
=
Q
A
N − 1
, D
0
=
Q
0
N − 1
, F =
nD
A
+ D
0
D
0
.
F
m−1, m(n −1)
A X
N
z = (S
1
+ S
2
+ S
3
)/15 = 7539, 0.
• Q
A
= z −
t
2
N
= 89, 2, D
A
= 2, 03
S
j
t
u
S
2
j
v
• Q
0
= u − z = 262, 0, D
0
= 5, 95
• Q = u −
t
2
N
= 351, 2
• F = 6, 12
Q
A
Q
n
p p
q q p
p
•
•
•
x
(1)
, ...x
(p)
X = (x
(1)
, ..., x
(p)
), Z = Z(X) q
q p Z(X) = (Z
(1)
(X), ..., Z
(q)
(X)) I
q
(Z(X))
I
F Z
˜
Z F
I
q
(
˜
Z(X)) = max
Z∈F
I
q
(Z(X)).
I, F
F
Z
(j)
=
p
X
k=1
c
j,k
(x
(k)
− ¯x
(k)
), j = 1, ..., q,
p
X
k=1
c
2
j.k
= 1, j = 1, ..., q,
p
X
k=1
c
j,k
c
i,k
= 0, j, k = 1, .., q, j 6= k.
I
q
(Z) =
P
q
j=1
DZ
(j)
P
p
j=1
Dx
(j)
.
x
(1)
, ..., x
(p)
S
1
, ..., S
q
Z
F
DZ = 1
I
q
(Z, S) =
q
X
j=1
X
k∈S
j
ρ
2
(x
(k)
, Z
(j)
),
ρ(., .) S
ρ
i,j
d
i,j
∆(Z) =
n
X
i=1
n
X
j=1
d
α
i,j
|
ˆ
d
i,j
(Z) − d
i,j
|
β
.
ˆ
d
i,j
α, β α < 0