Estimation and Sampling Distributions

8/12/2019 Estimation and Sampling Distributions

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Estimation and SamplingDistributions

Chapter 7


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Unbiasedness expected = true

Bias= = the difference betweenthe expected value of the estimator and the truevalue in the population.

Efficiency - Smallest Mean Squared Error

How well the estimator does in predicting.We want the estimator that has the smallestsquared error around the true value

Properties of Estimators that WeDesire

)

(E)

(E


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Efficiency is variance + squared bias

22

22

2

2

2

E

E

E

E

E

E2

E

E

E

E

E

E

E

E

E

E.E.S.M

Squared Bias

Variance

This is alwayszero


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Unbiasedness

BiasedUnbiased

P(X)

X


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Efficiency

SamplingDistributionof Median Sampling

Distribution ofMean

X

P(X)


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Largersample size

Smallersample size

Consistency

X

P(X)

A

B


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Estimation

Sample Statistic Estimates Population Parameter

e.g. X = 50 estimates Population Mean,

Problems: Many samples provide many estimates of thePopulation Parameter.

Determining adequate sample size: large sample give better

estimates. Large samples more costly.

How good is the estimate?

Approach to Solution: Theoretical Basis is SamplingDistribution.

_


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Sampling DistributionsSampling

Distributions

Sampling

Distributionsof theMean

Sampling

Distributionsof theProportion


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Sampling Distributions A sampling distribution is a

distribution of all of thepossible values of a statisticfor a given size sampleselected from a population


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Developing aSampling Distribution

Assume there is a population

Population size N=4Random variable, X,is age of individuals

Values of X: 18, 20,22, 24 (years)

A B CD


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.3

.2.1

0 18 20 22 24A B C D

Uniform Distribution

P(x)

x

(continued)

Summary Measures for the Population Distribution:


214

24222018N

X i

2.236N

)(X

2i


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1 st 2 nd ObservationObs 18 20 22 24

18 18,18 18,20 18,22 18,2420 20,18 20,20 20,22 20,24

22 22,18 22,20 22,22 22,24

24 24,18 24,20 24,22 24,2416 possible samples

(sampling withreplacement)

Now consider all possible samples of sizen=2

1st 2nd Observation

Obs 18 20 22 24

18 18 19 20 21

20 19 20 21 22

22 20 21 22 23

24 21 22 23 24

(continued)


16 SampleMeans


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1st 2nd Observation

Obs 18 20 22 24

18 18 19 20 21

20 19 20 21 22

22 20 21 22 23

24 21 22 23 24

Sampling Distribution of All SampleMeans

18 19 20 21 22 23 240

.1

.2

.3P(X)

X

Sample MeansDistribution

16 Sample Means

_


(continued)

(no longer uniform)

_


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Summary Measures of this SamplingDistribution:


(continued)

2116

24211918

N

X

i

X

1.5816

21)-(2421)-(1921)-(18N

)X(

222

2Xi

X


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Comparing the Population withits Sampling Distribution

18 19 20 21 22 23 240

.1

.2

.3P(X)

X18 20 22 24

A B C D

0

.1

.2

.3

PopulationN = 4

P(X)

X _

1.58 21 XX 2.236 21

Sample Means Distributionn = 2

_


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Estimation

Suppose that you want to know howmany tigers there are in the jungle.How could you use sampling to get agood estimate?


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Answer

Mark; release; resampleCatch 50 tigers. Put a band aroundtheir neck. Release them in the jungleagain. Now, catch 50 tigers again.What percentage are the originals that

were captured?How could this be used in otherestimations? On what does it rely?


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If the Population is NormalIf a population is normal with mean andstandard deviation , the sampling

distribution of is also normallydistributed with

and

X

X n

X


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Distributionof the Mean

Z-value for the sampling distribution of:

where: = sample mean= population mean= population standard deviationn = sample size

X

n

)X(

)X(Z X

X

X


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uNormal

We can apply the Central Limit Theorem :

Even if the population is not normal ,

sample means from the population will be approximately normal as long as the samplesize is large enough.

Properties of the sampling distribution:

and

x n

x


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Population Distribution

Sampling Distribution(becomes normal as n increases)

Central Tendency

Variation

(Sampling withreplacement)

x

x

Largersamplesize

Smallersample size

If the Population is not Normal (continued)

Sampling distributionproperties:

x

n

x

x


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How Large is Large Enough?

For most distributions, n > 30 willgive a sampling distribution that isnearly normal

For fairly symmetric distributions, n >

15For normal population distributions,the sampling distribution of the mean

is always normally distributed


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ExampleSolution:

Even if the population is not normally

distributed, the central limit theorem canbe used (n > 30)

so the sampling distribution of isapproximately normal

with mean = 8

and standard deviation

(continued)

x

x

0.5363

n

x


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ExampleSolution(continued):

(continued)

0.38300.5)ZP(-0.5

363

8-8.2

n

-

363

8-7.8P8.2)P(7.8 XX

Z7.8 8.2 -0.5 0.5

Sampling

Distribution

Standard Normal

Distribution .1915+.1915

Population

Distribution?

??

?

????

???? Sample Standardize

8 8 X 0 z xX


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Sampling Distributionsof the Proportion

SamplingDistributions

Sampling

Distributionsof theMean

Sampling

Distributionsof theProportion


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Sampling Distribution of p

Approximated by anormal distribution if:

where

and(where p = population proportion)

Sampling DistributionP( p s)

.3

.2

.10

0 . 2 .4 .6 8 1 p s

psp

np)p(1

sp

5p)n(1

5np

and


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Z-Value for Proportions

If sampling is withoutreplacement and n isgreater than 5% of thepopulation size, then

must use the finite

1NnN

np)p(1

sp

np)p(1

pp

ppZ s

p

s

s

Standardize p s to a Z value with the formula:

p


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Example

If the true proportion of voters whosupport Proposition A is p = .4, what isthe probability that a sample of size 200yields a sample proportion between .40

and .45?i.e.: if p = .4 and n = 200, what isP(.40 p s .45) ?


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Example if p = .4 and n = 200, what is

P(.40 p s .45) ?

(continued)

.03464200

.4).4(1n

p)p(1sp

1.44)ZP(0

.03464.40.45Z

.03464.40.40P.45)pP(.40 s

Find :

Convert tostandardnormal:

sp


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Example

Z.45 1.44

.4251Standardize

Sampling DistributionStandardized

Normal Distribution

if p = .4 and n = 200, what isP(.40 p s .45) ?

(continued)

Use standard normal table: P(0 Z 1.44) = .4251

.40 0p s

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Estimation and Sampling Distributions

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