n,

11.1368562.50%5

n p1-pnpn(1-p)5 ,

n50npn(1p)

un npn(1p)5 p

uu

1 20%()651524831.6%

u=3.58 > u0.05=1. 64, P

unpn(1-p)5

2 1

1

n1=100p1=14%n2=120p2=25%pc=20%1pc=80% u 2.031>u0.05=1.96p < 0.05=0.05H0H1

2

2 1.2.3.

2 1

1 2

2 41011-1

11-1 (%)4019023017.395013018027.789032041021.95

(50.49) (179.51) (39.51) (140.49)

2 2

2 P196 111

2 H02 22H0P

2 P

2 1.n40T5

1. H0 1=2 H1 1 2 =0.052. T11= 50.49 T12=179.51 T21= 39.51 T22=140.49

=(2-1)(2-1)=13.P =1 2 0.01

2.

Aaba+bBcdc+da+cb+da+b + c+d

2

3.11T5n402T1n40

11.6

11-2 % 13.76107.24119.091310.241719.763043.3314274134.15

X22.812X2 2.812 X2 0.05,1=3.84 P0.05X2 X2 =4.197P 0.05

3

3

3

X22.74 X2 0.05,1=3.84 P0.05X2 X2 =4.06P 0.05

2

211.7 55ITALAT RF 11-3

11-3 RF

ITALAT+-+31a12b43-1c11d12322355

H0 B=CH1 BC =0.05 =120.005

3 28+-4

4

1 H0B=C H1BC =0.05 22 b=9c=1b+c

3P X24.90X23.84P0.05=0.05H0H1

2RC =R-1C-1

nAnRnCA

=R-1C-1

11-4

%126997111.69126666781.77296656944.18357177524.6588274728353.10

H0 H1 =0.05

= 3 2 P < 0.005=0.05H0H1

11-5

753525123418129611869331466623711539445180843

H0 H1 =0.05

= 3 2 P > 0.05=0.05H0

1.1

1

X2

P =3121=2X20.05,25. 99=0.05H0H1

2.2

1

X2

P =2141=3 X2 0.05,3=7.81=0.05H0

1. 1/551

2.

3.