mendenhall chapter 10 esp

7/29/2019 Mendenhall Chapter 10 Esp

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Probabilidad y estadsticaProbabilidad y estadstica

Captulo 10

Inferencia con muestras

pequeas

Some graphic screen captures from Seeing Statistics Some images 2001-(current year) www.arttoday.com


2/41

IntroduccinIntroduccin Cuando el tamao de la muestra es

pequeo, la estimacin y los procedimientosde pruena del captulo 8 no son apropiados.

Existe un test de muestras pequeas equivalen-

te y procedimientos de estimacin para, la media de la poblacin normal12, la diferencia entre dos mediaspoblacionales

2, la varianza de una poblacin normalLa proporcin entre dos varianzas

poblacionales.


3/41

La distribucin muestral deLa distribucin muestral de

la media muestralla media muestral Cuando tomamos una muestra de una pobla-

cin normal, la media de la muestra tieneuna distribucin normal para cualquier tamao

n, y

Tiene distribucin normal estndar. Pero si es desconocida, y debemos estimarla,

el estadstico no es normalno es normal.

n

xz

/

= normal!esno

/ ns

x

x


4/41

Copyright 2006 Brooks/ColeA division of Thomson Learning, Inc.

Distribucin de StudentDistribucin de Student Afortunadamente, este estadstico posee una

distribucin muestral que es bien conocidapara los estadsticos, llamada distribucindistribucin

de Student,de Student, connn-1 grados de libertad.-1 grados de libertad.

ns

xt

/

=

Podemos utilizar esta distribucin para crear

procedimientos de prueba para la media de la

poblacin


5/41


Propiedades dePropiedades de ttde Studentde Student

La forma depende del tamao de la muestra

n o de los grados de libertad,grados de libertad, nn-1-1 A medida que n se incrementa, las formas

de las distribuciones tyzse tornan casi

idnticas.

Forma de monteForma de monte y

simtrica alrededor de 0

Ms variable queMs variable quezz, con

colas ms pesadas

APPLET

APPLETMY
http://var/www/apps/conversion/tmp/scratch_2/Beaver/studentT.htmlhttp://var/www/apps/conversion/tmp/scratch_2/Beaver/studentT.htmlhttp://var/www/apps/conversion/tmp/scratch_2/Beaver/studentT.html


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Usando la tablaUsando la tabla tt La tabla 4 da los valores de tque excluyen ciertos

valores crticos en la cola de la distribucin t Indexa dfdf y el rea apropiada de la cola aapara

encontrarttaa,, el valor de tcon rea aa su derecha.

Para una muestra al azar detamao n = 10, encuentre el

valor de tque deja 0,025 en la

cola derecha.

Fila = df= n 1 = 9

t0,025 = 2,262Subndice columna = a =0,025


7/41

Inferencia de una muestra chicaInferencia de una muestra chica

para la media poblacionalpara la media poblacional Los procesos bsicos son los mismos a los

utilizados para muestras grandes. Para untest de hiptesis:

1conndistribuciuna

enbasadarechazodereginunao-valoresusando

/

oestadstictestelutilizando

colasdosouna:Hversus:HPruebe

1

a11

=

=

=

ndft

p

ns

x

t


8/41


Para un intervalo de confianza de 100(1) % para

la media poblacional :

1conndistribuciunadecolalaen

/1reaundejaquedevalorelesdonde1/

1/

=

ndft

tt

n

stx

Inferencia de una muestra chicaInferencia de una muestra chica

para la media poblacionalpara la media poblacional


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EjemploEjemplo

Un sistema de aspersin est diseado de tal forma que eltiempo promedio de los picos para activarse luego dehaber sido encendidos es no mayor a 15 seg Una pruebade 5 sistemas di los tiempos siguientes:

17, 31, 12, 17, 13, 25Est trabajando el sistema como se especific? Pruebe

usando = 0,05

)especificsecomootrabajand(no11:H

)especificsecomoo(trabajand11:H

a

1

>=


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EjemploEjemploDatos:Datos: 17, 31, 12, 17, 13, 25

Primero calcule la media y la desviacinestndar muestral:

111,11

1

1111111

1

)(

111,1 11

111

111

=

=

=

==

=

n

n

xx

s

n

xx i


11/41


EjemploEjemploDatos:Datos: 17, 31, 12, 17, 13, 25

Calcule el test estadstico y encuentre laregin de rechazo para = 0,05

111111,11/1 1 1,111111,1 1

/

:libertaddeGrados:oestadsticTest

1 ====== ndfns

xt

Regin rechazo: Rechace H0 si t> 2,015 Si el test estadstico cae

en la regin de rechazo, su valor-

p debe ser menor a = 0,05

Regin rechazo: Rechace H0 si t

> 2,015 Si el test estadstico cae

en la regin de rechazo, su valor-

p debe ser menor a = 0,05


12/41


ConclusinConclusinDatos:Datos: 17, 31, 12, 17, 13, 25

Compare el test estadstico con la regin derechazo, y saque conclusiones.

111,1siHRechace

:rechazodeRegin

11,1:oestadsticTest

1>

=

t

t

Conclusin: Para nuestro ejemplo, t= 1,38 no cae en la reginde rechazo y H0 no es rechazada. La evidencia es insuficiente

para indicar que el tiempo promedio de activacin es mayor a

15.

11:H11

:Ha

1

>=


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Aproximando el valor-Aproximando el valor-pp Slo puede aproximar el valor-ppara la

prueba usando la Tabla 4.

Since the observed valueoft= 1.38 is smaller

than t.10 = 1.476,

p-value > .10.


14/41


El valor-El valor-pp exactoexacto

Puede obtener el valor-p usandoalgunas calculadoras o la PC.

One-Sample T: Times

Test of mu = 15 vs > 15

95%

Lower

Variable N Mean StDev SE Mean Bound T P

Times 6 19.1667 7.3869 3.0157 13.0899 1.38 0.113

One-Sample T: Times

Test of mu = 15 vs > 15

95%

Lower

Variable N Mean StDev SE Mean Bound T P

Times 6 19.1667 7.3869 3.0157 13.0899 1.38 0.113

Valor-p = .113 que es

mayor que .10 quehabamos aproximado

usando Tabla 4.

APPLETAPPLETMY
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Probando la diferenciaProbando la diferencia

entre dos mediasentre dos medias

normales.serdebenspoblacionedoslaspequeos,sonmuestraslasdetamaoslosqueDado

1

1y1

1sy varianza

1y1

mediascon1y1spoblacionede

extraense1

y1

tamaodeazaralntesindependiemuestraslas,1captuloelenComo

nn

Para probar:

H0: 12 = D0 versus Ha: una de tresdonde D0 es alguna diferencia que se ha

tomado como hiptesis, generalmente 0


16/41


Testing the DifferenceTesting the Difference

between Two Meansbetween Two MeansThe test statistic used in Chapter 9

does not have either azor a tdistribution, and

cannot be used for small-sample inference.We need to make one more assumption, thatthe population variances, although unknown,are equal.

1

1

1

1

1

1

11z

ns

ns

xx

+


17/41



between Two Meansbetween Two MeansInstead of estimating each population varianceseparately, we estimate the common variancewith

1

)1()1(

11

1

11

1

111

++

=nn

snsns

+

=

11

1

111

11

nns

Dxxt has a tdistribution

with n1+n2-2 degrees

of freedom.

And the resultingtest statistic,


18/41


Estimating the DifferenceEstimating the Difference

between Two Meansbetween Two MeansYou can also create a 100(1-)% confidenceinterval for1-2.

1

)1()1(with

11

1

11

1

111

++=

nn

snsns

+11

1

1/11

11)(

nnstxx

Remember the three

assumptions:

1. Original populations

normal

2. Samples random andindependent

3. Equal population

variances.

Remember the three

assumptions:1. Original populations

normal

2. Samples random and

independent

3. Equal population

variances.


19/41


ExampleExample Two training procedures are compared by

measuring the time that it takes trainees toassemble a device. A different group of trainees are

taught using each method. Is there a difference in the

two methods? Use = .01.

Time toAssemble

Method 1 Method 2

Sample size 10 12

Sample mean 35 31

Sample Std Dev 4.9 4.5

1:H111=

+

=

11

1

11

11

1

:statisticTest

nns

xxt

1:H11a


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ExampleExample Solve this problem by approximating thep-

value usingTable 4.

Time toAssemble

Method 1 Method 2

Sample size 10 12

Sample mean 35 31


11.1

1 1

1

11

11 1 1.11

111 1

:statisticTest

=

+

=t

111.1 11 1

)1.1(11)1.1(11

)1()1(

:Calculate

11

11

1

11

1

111

=+

=

+

+=

nn

snsns

APPLETAPPLETMY
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21/41


ExampleExample

value)-(1

1)11.1(

)11.1()11.1(:value-

ptP

tPtPp

=>

.025 < (p-value) < .05

.05


22/41



between Two Meansbetween Two MeansHow can you tell if the equal varianceassumption is reasonable?

statistic.testealternativanuse

,1smaller

largerratio,theIf

.reasonableisassumptionvarianceequalthe

,1smaller

largerratio,theIf

:ThumbofRule

1

1

1

1

>

s

s

s

s


23/41



between Two Meansbetween Two MeansIf the population variances cannot be assumedequal, the test statistic

has an approximate tdistribution with degreesof freedom given above. This is most easilydone by computer.

1

1

1

1

1

1

11

n

s

n

sxxt

+

1

)/(

1

)/(

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

+

+

n

ns

n

ns

ns

ns

df


24/41


The Paired-DifferenceThe Paired-Difference

TestTest

Sometimes the assumption of independentsamples is intentionally violated, resulting in amatched-pairsmatched-pairs orpaired-difference testpaired-difference test.

By designing the experiment in this way, we caneliminate unwanted variability in the experimentby analyzing only the differences,

ddii ==xx11ii xx22ii

to see if there is a difference in the twopopulation means, 12.

E l


25/41


ExampleExample

One Type A and one Type B tire are randomly assigned

to each of the rear wheels of five cars. Compare the

average tire wear for types A and B using a test of

hypothesis.

Car 1 2 3 4 5

Type A 10.6 9.8 12.3 9.7 8.8

Type B 10.2 9.4 11.8 9.1 8.3

1:H

1:H

11a

111

=

But the samples are not

independent. The pairs of

responses are linked becausemeasurements are taken on the

same car.


26/41


The Paired-DifferenceThe Paired-Difference

TestTest

.1on withdistributi-ta

onbasedregionrejectionaorvalue-theUse.s,differencetheofdeviationstandardandmean

theareandpairs,ofnumberwhere

/

1

statistictesttheusing

1:Htestwe1:HtestTo d1111

=

=

=

==

ndf

pd

sdn

nsdt

i

d

d

E l


27/41


ExampleExampleCar 1 2 3 4 5

Type A 10.6 9.8 12.3 9.7 8.8

Type B 10.2 9.4 11.8 9.1 8.3

Difference .4 .4 .5 .6 .5

1:H

1:H

11a

111

=

( )

.1111

.11Calculate

=

=

==

1

1

1

n

n

dd

s

n

dd

ii

d

i1.11

1/1111.

111.

/

1

:statisticTest

=

=

=ns

dt

d

E lE l


28/41


ExampleExampleCarCar 1 2 3 4 5

Type A 10.6 9.8 12.3 9.7 8.8

Type B 10.2 9.4 11.8 9.1 8.3

Difference .4 .4 .5 .6 .5

Rejection region: Reject H0

ift

> 2.776 ort< -2.776.

Conclusion: Since t= 12.8, H0

is rejected. There is a

difference in the average tirewear for the two types of tires.


29/41


Some NotesSome Notes

You can construct a 100(1-)% confidenceinterval for a paired experiment using

Once you have designed the experiment bypairing, you MUST analyze it as a pairedexperiment. If the experiment is not designed as apaired experiment in advance, do not use thisprocedure.

n

std d1/


30/41


Inference ConcerningInference Concerning

a Population Variancea Population VarianceSometimes the primary parameter of interestis not the population mean but rather thepopulation variance 2. We choose a randomsample of size n from a normal distribution.

The sample variances2 can be used in itsstandardized form:

1

1

1 )1(

sn =

which has a Chi-Squaredistribution with n - 1degrees of freedom.


31/41



a Population Variancea Population VarianceTable 5 gives both upper and lower criticalvalues of the chi-square statistic for a given df.

For example, the value ofchi-square that cuts off .

05 in the upper tail of the

distribution with df= 5 is

2 =11.07.


32/41



a Population Variancea Population Variance

.1on withdistributisquare-chia

onbasedregionrejectionawith)1(

statistictesttheusewe

tailedor twoone:Hversus:HtestTo

1

1

1

1

a

1

1

1

1

=

=

=

ndf

sn

1

)1

/1

(

1

1

1

1/

1 )1()1(

:intervalConfidence


33/41


ExampleExample

A cement manufacturer claims that his cementhas a compressive strength with a standarddeviation of 10 kg/cm2 or less. A sample ofn =

10 measurements produced a mean and standarddeviation of 312 and 13.96, respectively.

A test of hypothesis:

H0: 2 = 10 (claim is

correct)

Ha: 2 > 10 (claim is

wrong)

A test of hypothesis:

H0: 2 = 10 (claim iscorrect)

Ha: 2 > 10 (claim is

wrong)

uses the test statistic:uses the test statistic:

1.111 1 1

)11.1 1(111

)1(1

1

1

1 === sn


34/41


ExampleExample

Do these data produce sufficient evidence toreject the manufacturers claim? Use = .05.

Rejection region: Reject H0if2 > 16.919 ( = .05).

Conclusion: Since 2= 17.5,

H0 is rejected. The standard

deviation of the cement

strengths is more than 10.

A i i hA i ti th


35/41


Approximating theApproximating the

p-p-valuevalue

11with)1.1 1(:value- 1 ==> ndfPp

.025


36/41



Two Population VariancesTwo Population VariancesWe can make inferences about the ratio oftwo population variances in the form a ratio.We choose two independent random samplesof size n1and n2 from normal distributions.

If the two population variances are equal, thestatistic

1

1

1

1

s

s

F=

has anFdistribution with df1 = n1 - 1 and df2 =

n2 - 1 degrees of freedom.


37/41



Two Population VariancesTwo Population VariancesTable 6 gives only upper critical values of theF statistic for a given pair ofdf1and df2.

For example, the value of

F that cuts off .05 in the

upper tail of the

distribution with df1 = 5and df2 = 8 is F =3.69.


38/41



Two Population VariancesTwo Population Variances

.1and1

on withdistributianonbasedregionrejectionawith

.variancessampletwotheoflargertheiswhere

statistictesttheusewe

tailedor twoone:Hversus:HtestTo

1111

1

11

1

1

1

a

1

1

1

11

==

=

=

ndfndf

F

ss

s

F

11

11

,1

1

11

1

1

11

,

1

1

11 1

:intervalConfidence

dfdf

dfdf

Fs

s

Fs

s


39/41


ExampleExample

An experimenter has performed a labexperiment using two groups of rats. He wantsto test H0: 1 = 2, but first he wants to make

sure that the population variances are equal.Standard (2) Experimental (1)

Sample size 10 11

Sample mean 13.64 12.42


1

1

1

1a

1

1

1

11:Hversus:H

:y testPreliminar

=


40/41


ExampleExampleStandard (2) Experimental (1)

Sample size 10 11


1

1

1

1a

1

1

1

11

:H

:H

=

11.11.1

1.1

:statisticTest

1

1

1

1

11 ===

s

sF

We designate the sample with the larger standarddeviation as sample 1, to force the test statistic

into the upper tail of theFdistribution.

We designate the sample with the larger standarddeviation as sample 1, to force the test statistic

into the upper tail of theFdistribution.


41/41

C i h 2006 B k /C l

ExampleExample

1

1

1

1a

1

1

1

11

:H:H

= 11.11.1

1.1

:statisticTest

1

1

1

1

1

1 ===s

sF

The rejection region is two-tailed, with = .05, but we onlyneed to find the upper critical value, which has /2 = .025 to

its right.

From Table 6, with df1=10 and df2 = 9, we reject H0 ifF>

3.96.

CONCLUSION: Reject H0. There is sufficient evidence to

indicate that the variances are unequal. Do not rely on the

i f l i f !

The rejection region is two-tailed, with = .05, but we onlyneed to find the upper critical value, which has /2 = .025 to

its right.

From Table 6, with df1=10 and df2 = 9, we reject H0 ifF>

3.96.

CONCLUSION: Reject H0. There is sufficient evidence to

indicate that the variances are unequal. Do not rely on the

assumption of equal variances for yourt test!

mendenhall chapter 10 esp

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