9 Статистические методы иследования пола
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101
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2 (5 7,1)2 (12 8,6)2 (8 9,3)2 2,14
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1=50 ( ) |
2=250 ( ) |
1 2=300 |
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3=10 ( ) |
4=190 ( ) |
3 4=200 |
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1 3=60 |
2 4=440 |
n=500 |
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. . 5. .
. – : , 2006. –208 .
102
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2 .
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25 |
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( 1+ 2)·( 1+ 3):n |
( 2+ 1)·( 2+ 4):n |
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(50+250)·(50+10):500 |
(250+50)·(250+190):500 |
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1 =36 ( ) |
2 =264 ( ) |
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( 3+ 1)·( 3+ 4):n |
( 4+ 2)·( 4+ 3):n |
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(10+50)·(10+190):500 |
(50+250)·(50+10):500 |
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3 =24 ( ) |
4 =176 ( ) |
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2 ( )2 .
2 (50 36)2 (250 264)2 (10 24)2 (190 176)2
36 |
264 |
24 |
176 |
=5,44+0,74+8,17+1,11 = 15,46
v=(lx–1)·(ly–l) =(2–1) · (2–1)=1. . 34 ( . 124)
2. =1 =0,01 6,63. -
2 (15,46) (6,63), ,
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. . 5. .
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103
9.4.6.3.2
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3=( 1 2):2 |
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(5+9) :2=7 |
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1,5 |
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(6+10) |
:2=8 |
1,8 |
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16 |
(8+16) |
:2=12 |
2,1 |
10 |
20 |
(10+20) :2=15 |
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2,4 |
20 |
16 |
(20+16) :2=18 |
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2,7 |
14 |
14 |
(14+14) :2=14 |
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3,0 |
17 |
5 |
(17+5) |
:2=11 |
3,3 |
15 |
5 |
(15+5) |
:2=10 |
3,6 |
5 |
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(5+5) :2=5 |
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1=100 |
2=100 |
3=100 |
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. . 5. .
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104
2
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3=( 1 2):2. ( . 27).
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–22:7=0,57 |
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+22:7=0,57 |
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–2 |
–22:8=0,50 |
10 |
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+2 |
+22:8=0,50 |
8 |
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–4 |
–42:12=1,33 |
16 |
12 |
+4 |
+42:12=1,33 |
10 |
15 |
–5 |
–52:15=1,66 |
20 |
15 |
+5 |
+52:15=1,66 |
20 |
18 |
+2 |
+22:23=0,17 |
16 |
18 |
–2 |
–22:23=0,17 |
14 |
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02:14=0 |
14 |
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02:14=0 |
17 |
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+62:11=3,27 |
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–62:11=3,27 |
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+52:10=2,50 |
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–52:10=2,50 |
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02:5=0 |
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02:5=0 |
(p1–p2)= 200 |
2p3=200 |
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2 = 2 =20 |
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. . 5. .
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105
2
2 =20.
=n–1=9–1=8. (34 . 124)
2 =8. : ) 15,5 =0,95, ) 20,1
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( . 24) 2, -
(38), :
2 (50 190 250 10)2 100 (9500 2500)2 500 24500 15,45
300 200 60 440 |
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HaHa, HaH–– |
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b=95 |
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H––H–– |
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d=175 |
+ d = 190 |
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b + d = 270 |
n=a+b+c+d=310 |
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. . 5. .
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106
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(36)
v = (2 – 1)·(2 – 1) = 1.
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. . 5. .
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107
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1)(« », « » .)
( 1·n2). ;
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1/n1·n2 ( . 39).
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. . 5. .
. – : , 2006. –208 .
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(33390 29878 60975 11805) |
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3,24 |
200 210 |
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2=3,24; 5) = l–1 = 4–l =3.
=0,05 2=7,81 2
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. . 5. .
. – : , 2006. –208 .
109
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. . 5. .
. – : , 2006. –208 .
1. , -
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1.3.x
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1.4.?
1.5.?
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1.7.?
1.8..
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3.?
4.-
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4.1. -
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6.1.2?
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6.3.2?
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6.5.2 -
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6.6.2
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