UNIVERSIDAD

GUADALAJARA LAMAR

Titulo: Proyecto final

Investigador:

Neri Carrillo Alejandra

Asesor: Bejar Rivera Ana Leticia

Materia: Bioestadisitica inferencial

Carrera: Medicina 2A

09/12/15

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

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

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 29 29

19

13

5

3

1 0 0

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

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 �̅�

�̅� =𝟓𝟕. 𝟒𝟓

𝟐𝟗= 𝟏. 𝟗𝟖

�̅� = 𝟎. 𝟖𝟓 +𝟑𝟐. 𝟖

𝟐𝟗= 𝟏. 𝟗𝟖

Mediana 𝑥

𝑥 = 𝟎. 𝟒𝟓 + (

𝟐𝟗𝟐 − 𝟎

𝟏𝟎)𝟎. 𝟖 = 𝟏. 𝟔𝟏

𝑥 = 𝟏. 𝟐𝟓 − (

𝟐𝟗𝟐 − 𝟏𝟗

𝟏𝟎)𝟎. 𝟖 = 𝟏. 𝟔𝟏

Moda 𝑥

𝑥 = 𝟎. 𝟒𝟓 + (𝟏𝟎

𝟏𝟒) = 𝟏. 𝟎𝟐

𝑥 = 𝟏. 𝟐𝟓 − (𝟒

𝟏𝟒) = 𝟏. 𝟎𝟐

MEDIDAS DE DISPERCIÒN O VARIACIÒN

| x – 𝒙 | | x – 𝒙 |2 f| x – �̅� | f| x – �̅� |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. 𝟏𝟎𝟐𝟐 = 𝟏. 𝟐𝟏𝟒

Coeficiente de variación (CV)

𝐶𝑉 =𝟏. 𝟏𝟎𝟐

𝟏. 𝟗𝟖(𝟏𝟎𝟎) = 𝟓𝟓. 𝟔𝟓%

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

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

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:

�̅� = 1.98

δ = 1.36

n = 10

N = 29

δ𝑥 = (1.36

√10)(√

29 − 10

29 − 1) = 𝟎. 𝟏𝟏𝟏

Al 90% ---> 1.644

�̅� ± (Ζ𝒸)(δ𝑥) = 1.98 + (1.644)(0.111) = 𝑁𝐶𝑆 2.162 Margen

0.365 1.98 − (1.644)(0.111) = 𝑁𝐶𝐼 1.797

Al 99% ---> 2.588

�̅� ± (Ζ𝒸)(δ𝑥) = 1.98 + (2.588)(0.111) = 𝑁𝐶𝑆 2.267 Margen

0.575 1.98 − (2.588)(0.111) = 𝑁𝐶𝑆 1.692

𝒕 -STUDENT

Datos:

�̅� = 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

�̅� ± (t𝒸) (δ

√𝑁 − 1)

1.98̅̅ ̅̅ ̅̅ + (1.70) (1.36

√29 − 1) = 𝐿𝑆 2.416

Margen

0.873 1.98̅̅ ̅̅ ̅̅ − (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

�̅� ± (t𝒸) (δ

√𝑁 − 1)

1.98̅̅ ̅̅ ̅̅ + (2.76) (1.36

√29 − 1) = 𝐿𝑆 2.689

Margen

1.419 1.98̅̅ ̅̅ ̅̅ − (2.76) (

1.36

√29 − 1) = 𝐿𝐼 1.270

HIPOTESIS

Datos:

μ = 2.85

𝒩 = 29

�̅� = 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

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%

05

101520253035404550556065707580859095

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

CAMPANA DE GAUSS

𝜁 =𝐿𝑅𝐼 − �̅�

𝑆

�̅� = 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