61. (Point estimation)
2. (Interval estimation)
P(a < ( < b) = 1 - (
a b (Confidence limits)
(a , b) (Confidence interval)
( 1 - ( ) (Confidence coefficient)
(a , b) ( 1 - ( ) 100%
(
)
(
)
(
)
( 1 - ( ) 100% (
0.025
2
a
22
( n < 30) ( t ( n – 1 )
( 1 - ( ) 100% (
ˆˆ
pq
pq
n
X
X(N(( , 152)
( 1 - ( ) 100% = 90%

2
2
a
c
400 (n ( 30) (2 (2 S2 S2= (9.9) 2 Z0.025= 1.96 (
2
2
1
a
3 20 20
55 62 37 66 74 50 58 57 64 61
34 40 58 60 59 63 70 30 24 56
95%

(
)
(
)
(
)
(
)
(
)
(
)
0

2.93.1
0.158
21
0.64 - 0.030 < P < 0.64 + 0.030
0.61< P < 0.67
2
ZZ
a
,(1)
2
n
tt
a
14 147.6 26.384 95%
S = 26.384 , S2= 696.1 , n = 14 , df = (n – 1) = 13
1.96
Z
(Hypothesis testing)
H1 : (1 > (2
H1 : (1 < (2
H1 : p > 0.90
H1 : p < 0.90
(2

P( H0 H0 ) = (
2. 2 (Type II error) H0 H0 2 (
P( H0 H0 ) = (
(1 - () H0 H0 (Power of the test) ( (1 - () H0 H0 (1 - () 100 (1 - ()100% (Confidence level)


H1 : ( > (0 H1 : ( < (0
1 H1 : ( > (0
2 H1 : ( < (0
2. (Two – tailed test Two – side test)
H0 : ( = (0
H1 : ( ( (0
2. (
1
H0 : ( = (0 (
0.05
2
1.645
ZZ
a
t (n-1) t ( t( n – 1 )
H0 : ( = (0 (
1
8.6 0.3 36 8.7 .05
1.
ˆ
p
4.
ˆ
ˆˆ
pp
Z
pq
n
0.05
s2
t =
3. ( = n – 1 = 21 – 1 = 20 ( = 0.05 t < -1.725
4.

p0 . .
p0 = 0.80 , n = 200 , x = 136
68
4.


1.
(
)
5
2. 15
49 59 68 65 57 56 46 68 70 38 65 54 72 55 63
99%
3. 64 8.25 3 95%
5. 8 16 6.63 2.55 8 0.05