Proyecto final de bioestadistica Alumna Alejandra Neri Carrillo Universidad Guadalajara LAMAR Lic. Ana Leticia Bejar Rivera

1. UNIVERSIDAD
GUADALAJARA LAMAR
Titulo: Proyecto final
Investigador:
Neri Carrillo Alejandra
Asesor: Bejar Rivera Ana Leticia
Materia: Bioestadisitica inferencial
Carrera: Medicina 2A
09/12/15

2. ESTADISTICA DESCRIPTIVA
Distribución porcentual de mastografías realizadas en instituciones públicas
de salud, por entidad federativa, 2013.
0.5 0.9 1.3 1.9 2.3 3.6
N = 29
R = 4
K = 5
IC =0.8
0.7 1.0 1.4 2.1 2.6 3.9
0.7 1.0 1.7 2.1 2.7 4.0
0.8 1.1 1.8 2.2 2.7 4.9
0.9 1.1 1.9 2.2 3.5
K LI - LS f X LRI - LRS C fr Gráfica circular
1 0.5 – 1.2 10 0.85 0.45 – 1.25 0.8 34.48% 124º
2 1.3 – 2.0 6 1.65 1.25 – 2.05 0.8 20.68% 74º
3 2.1 – 2.8 8 2.45 2.05 – 2.85 0.8 27.58% 99º
4 2.9 – 3.6 2 3.25 2.85 – 3.65 0.8 6.89% 25º
5 3.7 – 4.4 2 4.05 3.65 – 4.45 0.8 6.89% 25º
6 4.5 – 5.2 1 4.85 4.45 – 5.25 0.8 3.44% 12º
29 99.96% 359º

3. 34.48%
20.68%
27.58%
6.89%
6.89%
3.44%
Grafica Circular
0
10
6
8
2 2
1
0
0
2
4
6
8
10
12
0.05 0.85 1.65 2.45 3.25 4.05 4.85 5.65
f
x
Histograma
𝑥 = 1.09

4. FRECUENCIA ACUMULADA
Menor LRI & LRS Mayor
0 0.45 29
10 1.25 19
16 2.05 13
24 2.85 5
26 3.65 3
28 4.45 1
29 5.25 0
0
10
16
24
26
28
2929
19
13
5
3
1
00
5
10
15
20
25
30
35
0.45 1.25 2.05 2.85 3.65 4.45 5.25
fac
LRI & S
Ojiva
Menor
Mayor
𝑥 = 1.89

5. MEDIDAS DE TENDENCIA CENTRAL
f * x d = x - A f * d
8.5 0 0
9.9 0.8 4.8
19.6 1.6 12.8
6.5 2.4 4.8
8.1 3.2 6.4
4.85 4 4
57.45 32.8
Media 𝑥̅
𝑥̅ =
𝟓𝟕. 𝟒𝟓
𝟐𝟗
= 𝟏. 𝟗𝟖
𝑥̅ = 𝟎. 𝟖𝟓 +
𝟑𝟐. 𝟖
𝟐𝟗
= 𝟏. 𝟗𝟖
Mediana 𝑥
𝑥 = 𝟎. 𝟒𝟓 + (
𝟐𝟗
𝟐
− 𝟎
𝟏𝟎
) 𝟎. 𝟖 = 𝟏. 𝟔𝟏
𝑥 = 𝟏. 𝟐𝟓 − (
𝟐𝟗
𝟐
− 𝟏𝟗
𝟏𝟎
) 𝟎. 𝟖 = 𝟏. 𝟔𝟏
Moda 𝑥
𝑥 = 𝟎. 𝟒𝟓 + (
𝟏𝟎
𝟏𝟒
) = 𝟏. 𝟎𝟐
𝑥 = 𝟏. 𝟐𝟓 − (
𝟒
𝟏𝟒
) = 𝟏. 𝟎𝟐

6. MEDIDAS DE DISPERCIÒN O VARIACIÒN
| x – 𝒙̅ | | x – 𝒙̅ |2 f| x – 𝒙̅ | f| x – 𝒙̅ |2 f * x2
1.13 1.276 11.3 12.76 7.225
0.33 0.108 1.98 0.648 16.335
0.47 0.220 3.76 1.76 48.02
1.27 1.612 2.54 3.224 21.125
2.07 4.284 4.14 8.568 32.805
2.87 8.236 2.87 8.236 23.522
26.59 35.196 149.032
Desviación media (DM)
𝐷𝑀 =
𝟐𝟔. 𝟓𝟗
𝟐𝟗
= 𝟎. 𝟗𝟏𝟔
Desviación estándar (S)
S = √
𝟑𝟓. 𝟏𝟗𝟔
𝟐𝟗
= 𝟏. 𝟏𝟎𝟏
S = √((
𝟏𝟒𝟗. 𝟎𝟑𝟐
𝟐𝟗
) − (
𝟓𝟕. 𝟒𝟓
𝟐𝟗
) 𝟐) = 𝟏. 𝟏𝟎𝟐
Variación (V)
V = 𝟏. 𝟏𝟎𝟐 𝟐
= 𝟏. 𝟐𝟏𝟒
Coeficiente de variación (CV)
𝐶𝑉 =
𝟏. 𝟏𝟎𝟐
𝟏. 𝟗𝟖
(𝟏𝟎𝟎) = 𝟓𝟓. 𝟔𝟓%

7. TEORIA DEL MUESTREO
Estratificado
Estratos Mastografias realizadas en
instituciones publicas
Cantidad fr Muestra
10 0.3448
6 0.2068
8 0.2758
2 0.0689
2 0.0689
1 0.0344
N= 29 0.9996

8. DISTRIBUCION DE MUESTREO DE MEDIAS
A) μ =
𝟎.𝟖𝟓+𝟏.𝟔𝟓+𝟐.𝟒𝟓+𝟑.𝟐𝟓+𝟒.𝟎𝟓+𝟒.𝟖𝟓
𝟔
=
𝟏𝟕.𝟏
𝟔
= 𝝁 = 𝟐. 𝟖𝟓
B) δ2
=
𝟏𝟏.𝟐
𝟔
= 𝟏. 𝟖𝟔
δ = √𝟏. 𝟖𝟔 = 𝟏. 𝟑𝟔
C)
102.6
36
= 𝟐. 𝟖𝟓
D) =
42.75
15
= 𝟐. 𝟖𝟓
δ𝑥2
=
11.2
15
= 𝟎. 𝟕𝟒
δx = √0.74 = 𝟎. 𝟖𝟔
Error estándar
δ2
= (
1.86
2
) (
6 − 2
6 − 1
) = (0.93)(0.8) = 0.744
0.85 1.65 2.45 3.25 4.05 4.85 Total
0.85
0.85,0.85
0.85
0.85,1.65
1.25
0.85,2.45
1.65
0.85,3.25
2.05
0.85,4.05
2.45
0.85,4.85
2.85 =11.1
1.65
1.65,0.85
1.25
1.65,1.65
1.65
1.65,2.45
2.05
1.65,3.25
2.45
1.65,4.05
2.85
1.65,4.85
3.25 =13.5
2.45
2.45,0.85
1.65
2.45,1.65
2.05
2.45,2.45
2.45
2.45,3.25
2.85
2.45,4.05
3.25
2.45,4.85
3.65 =15.9
3.25
3.25,0.85
2.05
3.25,1.65
2.45
3.25,2.45
2.85
3.25,3.25
3.25
3.25,4.05
3.65
3.25,4.85
4.05 =18.3
4.05
4.05,0.85
2.45
4.05,1.65
2.85
4.05,2.45
3.25
4.05,3.25
3.65
4.05,4.05
4.05
4.05,4.85
4.45 =20.7
4.85
4.85,0.85
2.85
4.85,1.65
3.25
4.85,2.45
3.65
4.85,3.25
4.05
4.85,4.05
4.45
4.85,4.85
4.85 =23.1
Total =102.6

9. ESTADISTICA INFERENCIAL
NIVEL DE CONFIANZA
Nivel de
confianza 99.75% 99% 98% 96% 95.45% 95% 90% 80% 60% 50%
ZC 3.000 2.588 2.333 2.055 2.000 1.966 1.644 1.288 1.000 0.677
Datos:
𝑥̅ = 1.98
δ = 1.36
n = 10
N = 29
δ𝑥 = (
1.36
√10
) (√
29 − 10
29 − 1
) = 𝟎. 𝟏𝟏𝟏
Al 90% ---> 1.644
𝑥̅ ± (Ζ𝒸)(δ𝑥) =
1.98 + (1.644)(0.111) = 𝑁𝐶𝑆 2.162 Margen
0.3651.98 − (1.644)(0.111) = 𝑁𝐶𝐼 1.797
Al 99% ---> 2.588
𝑥̅ ± (Ζ𝒸)(δ𝑥) =
1.98 + (2.588)(0.111) = 𝑁𝐶𝑆 2.267 Margen
0.5751.98 − (2.588)(0.111) = 𝑁𝐶𝑆 1.692

10. 𝒕 -STUDENT
Datos:
𝑥̅ = 1.98
δ = 1.36
n = 10
N = 29
Al 90% ---> 0.05
100% − 90% =
10
100
=
0.1
2
= 0.05
𝒕 = 1 − 0.05 = 0.95
𝒕 = 0.95
±1.70
𝒗 = 28
𝑥̅ ± (t𝒸) (
δ
√𝑁 − 1
)
1.98̅̅̅̅̅̅ + (1.70) (
1.36
√29 − 1
) = 𝐿𝑆 2.416
Margen
0.873
1.98̅̅̅̅̅̅ − (1.70) (
1.36
√29 − 1
) = 𝐿𝐼 1.543
Al 99% ---> 0.005
100% − 99% =
1
100
=
0.01
2
= 0.005
𝒕 = 1 − 0.005 = 0.995
𝒕 = 0.995
±2.76
𝒗 = 28
𝑥̅ ± (t𝒸) (
δ
√𝑁 − 1
)
1.98̅̅̅̅̅̅ + (2.76) (
1.36
√29 − 1
) = 𝐿𝑆 2.689
Margen
1.419
1.98̅̅̅̅̅̅ − (2.76) (
1.36
√29 − 1
) = 𝐿𝐼 1.270

11. HIPOTESIS
Datos:
μ = 2.85
𝒩 = 29
𝑥̅ = 1.98
δ = 1.36
𝐻0 = ℳ = 2.85
𝐻1 = ℳ ≠ 2.85
𝑡 =
1.98 − 2.85
1.36
(√29 − 1) = 𝟑. 𝟑𝟖
Al 90%
100% − 90% =
10
100
=
0.1
2
= 0.05
𝒕𝒄 = 1 − 0.05 = 0.95
𝒕𝒄 = 0.95
±1.70 RECHAZADA
𝒗 = 28
Al 99%
100% − 99% =
1
100
=
0.01
2
= 0.005
𝒕 = 1 − 0.005 = 0.995
𝒕 = 0.995
±2.76 RECHAZADA
𝒗 = 28

12. CUARTILES, DECILES Y PERCENTILES
Cuartiles
𝑞2 =
(2)29
4
= 14.5 𝑄1 = 1.25 + (
14.5 − 10
6
) 0.8 = 1.85
𝑞3 =
(3)29
4
= 21.75 𝑄3 = 2.05 + (
21.75 − 16
8
) 0.8 = 2.62
𝑞4 =
(4)29
4
= 29 𝑄4 = 5.25 + (
29 − 29
29
) 0.8 = 5.25
Deciles
𝑑4 =
(4)29
10
= 11.6 𝐷4 = 1.25 + (
11.6 − 10
6
) 0.8 = 1.46
𝑑7 =
(7)29
10
= 20.3 𝐷7 = 2.05 + (
20.3 − 16
8
) 0.8 = 2.48
𝑑9 =
(9)29
10
= 26.1 𝐷9 = 3.65 + (
26.1 − 26
2
) 0.8 = 3.69
Percentiles
𝑝13 =
(13)29
100
= 3.77 𝑃13 = 0.45 + (
3.77 − 0
10
) 0.8 = 0.75
𝑝24 =
(24)29
100
= 6.96 𝑃24 = 0.45 + (
6.96 − 0
10
) 0.8 = 1.00
𝑝48 =
(48)29
100
= 13.92 𝑃48 = 1.25 + (
13.92 − 10
6
) 0.8 = 1.77
Fac %
0 0%
10 34.48%
16 55.17%
24 82.75%
26 89.65%
28 96.55%
29 100%

13. 0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
0.45 1.25 2.05 2.85 3.65 4.45 5.25
Q2
Q3
D4
D7
D9
P13
P24
P48
Q4
% fac
LRI

14. CAMPANA DE GAUSS
𝜁 =
𝐿𝑅𝐼 − 𝑥̅
𝑆
𝑥̅ = 1.98
S = 1.10
K f LRI 𝜻 Area de la tabla Area de trabajo f
SI SE AJUSTA
A LA
CAMPANA
DE GAUSS
1 10 0.45 -1.39 0.4177 0.1723 5 NO
2 6 1.25 -0.66 0.2454 0.2693 8 SI
3 8 2.05 0.06 0.0239 0.2614 8 SI
4 2 2.85 0.79 0.2852 0.1493 4 SI
5 2 3.65 1.51 0.4345 0.053 2 SI
6 1 4.45 2.24 0.4875 0.011 1 SI
33 5.25 2.97 0.4985