Statistics">
Ejercicios Del Tamao de La Muestra Compress
Ejercicios Del Tamao de La Muestra Compress
Ejercicios Del Tamao de La Muestra Compress
N=10000
σ =3000 Z 2 2 N
n
α=0.05 ( N 1) 2 Z 2 2
e=500
z=1.96
345744000000
n=
2534324400
n= 136.42
N=8 000
σ =0.36 Z 2 2 N
n
α=0.05 ( N 1) 2 Z 2 2
e=?
z=1.96
n=300
300
= 0.0375 < 0.1
8000
8000 ∗ 0.362 ∗ 1.962
300 =
(8000 − 1) ∗ e2 + 1.962 ∗ 0.362
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 = 𝟒. 𝟗𝟗𝟓
93. ¿Qué tamaño deberá tener una muestra para estimar dentro del 3%, la
proporción de mujeres casadas que van periódicamente a consulta
ginecológica, en una población de 5000 mujeres y una seguridad del 95%?
N=5 000
P=0.5
Q=0.5 Z 2 PQN
n
α=0.05 (N 1) 2 Z 2 PQ
e=0.03
z=1.96
4802
𝑛= = 879.568
5.4595
N=12000
σ =1800 Z 2 2 N
n
α=0.01 ( N 1) 2 Z 2 2
e=-300 y 300
z=2.97
2.972 ∗18002 ∗12000
n = (12000−1)∗3002+2.972∗18002
342956592000
n=
28579716
n= 309.390864930677
Respuesta : 309 estudiantes
N= Tamaño de la población
Z = Nivel de confianza
P= Probabilidad de éxito.
Q = Probabilidad de fracaso
D = Precisión
70∗(0,952 )∗0,05∗0,35
n=
0,012 ∗(0,952 )∗0,05∗0,35
1,1055
n= = 70,41
0,0157
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂: 𝟕𝟎 𝒔𝒆𝒓í𝒂 𝒍𝒂 𝒎𝒖𝒆𝒔𝒕𝒓𝒂 𝒓𝒆𝒄𝒐𝒎𝒆𝒏𝒅𝒂𝒅𝒂
e=-0.03 Z 2 PQ
n
z=-2.01 2
α=0.05
P=0.6
Q=0.4
−2.012 ∗0.6∗0.4
n= 0.032
0.96934
n=
0.0009
n = 1077.36 =
Z 2 PQ
n
e=-0.03 2
z=-2.01
α=0.045
P=0.5
Q=0.5
−2.012 ∗0.5∗0.5
n= 0.032
1.010025
n= 0.0009
n = 1122.25 =
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 ∶ 𝟏𝟏𝟐𝟐 𝒇𝒂𝒎𝒊𝒍𝒊𝒂𝒔
c)
e=-0.03 Z 2 PQ
n
z=-2.01 2
α=0.045
P=0.9
Q=0.1
−2.012 ∗0.9∗0.1
n=
0.032
0.363609
n=
0.0009
n = 404.01 =
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 ∶ 𝟒𝟎𝟒 𝒇𝒂𝒎𝒊𝒍𝒊𝒂𝒔
N=10 000
P=0.6
2
Q=0.4 Z PQN
n
α=0.045 (N 1) 2 Z 2 PQ
e=0.03
z=-2.01
10100.25
𝑛= = 1009.104
10.009125
𝑛 = 1009 𝑓𝑎𝑚𝑖𝑙𝑖𝑎𝑠
1009
𝑛= = 0.1009 > 0.1
10000
n
𝑛= n = 916.52 = 916 𝑓𝑎𝑚𝑖𝑙𝑖𝑎𝑠
1+𝑁
N=20 000
P=0.85
Z 2PQN
n
Q=0.15 ( N 1) 2 Z 2 PQ
α=0.025
e=0.03
z=-2.24
10100.25
𝑛= = 1014.13
9.9595
𝑛 = 1014 𝑓𝑎𝑚𝑖𝑙𝑖𝑎𝑠
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 ∶ 𝟏𝟎𝟏𝟒 𝒇𝒂𝒎𝒊𝒍𝒊𝒂𝒔
𝑍² 𝑃 𝑄
𝑛=
𝛿²
(2.57)2 (0.2)(0.8)
𝑛=
(0.07)²
𝑛 = 215.67 ≅ 216 𝑝𝑒𝑟𝑠𝑜𝑛𝑎𝑠
N=365
σ =12
Z 2 2 N
α=0.1 n
e=12 ( N 1) 2 Z 2 2
z=1.69
150116.616
n=
52827.2784
n= 2.84
n=3 días
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 = 𝟑 𝒅í𝒂𝒔
3
n= = 0.008 < 0.1
365
σ=2800
α=0.045
e=500
z=-2
(2800∗(−2))2
n= 5002
31360000
n=
250000
n = 129.44
n = 129 𝑓𝑎𝑚𝑖𝑙𝑖𝑎𝑠
e= 0.04
z= -2 2
Z PQN
α= 0.045 n
(N 1) 2 Z 2 PQ
P= 0.5
Q= 0.5
N=3200
3200 ∗ 0.5 ∗ 0.5 ∗ −22
𝑛=
(3200 − 1) ∗ 0.042 + −22 ∗ 0.5 ∗ 0.5
3200
n=
6.12
n = 522.88
n = 523 alumnos
523
n= = 0.16> 0.10
3200
n
𝑛= n = 450.86 = 451 𝑎𝑙𝑢𝑚𝑛𝑜𝑠
1+𝑁
𝑍 2 𝑃𝑄𝑁
𝑛=
(𝑁 − 1) 𝛿² + 𝑍 2 𝑃 𝑄
(2.57)2 (0.5)(0.5)(10000)
𝑛=
(10000 − 1) (0.02)² + (2.57)2 (0.5)(0.5)
16512.25
𝑛=
5.650825
𝑛 = 2922.0954 ≅ 2922
Pero:
𝑛 2922
= = 0.2922 > 0.10
𝑁 10000
Entonces:
𝑛 2922
𝑛𝑐 = 𝑛 → 𝑛𝑐 =
1+ 𝑁 2922
1 + 10000
𝑛𝑐 = 2261.2599 ≅ 2261
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 ≅ 𝟐𝟐𝟔𝟏
103.
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 = 𝟑𝟓
(1.64)2 (0.3)(0.7)(30)
𝑛=
(30 − 1) (0.04)² + (1.64)2 (0.3)(0.7)
16.94448
𝑛=
0.611216
𝑛= 27.7226 ≅ 28 ℎ𝑜𝑔𝑎𝑟𝑒𝑠
Pero:
𝑛 28
= = 0.933 > 0.10
𝑁 30
Entonces:
𝑛 28
𝑛𝑐 = 𝑛 → 𝑛𝑐 =
1+ 28
𝑁 1 + 30
𝑛𝑐 = 14.483 ≅ 14 ℎ𝑜𝑔𝑎𝑟𝑒𝑠
𝑹𝒆𝒔𝒑𝒖𝒆𝒔𝒕𝒂 =≅ 𝟏𝟒 𝒉𝒐𝒈𝒂𝒓𝒆𝒔
N=12 500 Z 2 2 N
n
σ=3000 ( N 1) 2 Z 2 2
α=0.045
e=300
z=-2
−22 ∗30002 ∗12500
n= (12500−1)∗3002 +22 ∗30002
450000000000
n= = 387.63
1160910000
n= 387 familias
387
n= = 0.031< 0.10
12500