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llpp ggnniimmaaSSMethodsMethods&&

Central Limit TheoremCentral Limit Theorem

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We use sample informationtomake decisions or inferences

about the population.

We use sample informationtomake decisions or inferences

about the population.

TwoKEYsteps:TwoKEYsteps:

1. Choice of a proper method for selecting sample data

&2. Proper analysis of the sample data (more later)

KEY 1.KEY 1.

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KEY 1. KEY 1.If the proper

method for selecting

the sample is

NOT MADE

If the proper

method for selecting

the sample is

NOT MADE the SAMPLEwill not be truly

representative of theTOTAL Population!

the SAMPLEwill not be truly

representative of theTOTAL Population!

and wrong conclusions can be drawn!

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of the physical impossibility of checkingall items in the population, and,

also, it would be too time-consuming

$ the studying of all the items in a populationwould NOT be cost effective

the sample results are usually sufficient

the destructive nature of certain tests

Why Sample the Population?Why Sample the Population?Why Sample the Population?Why Sample the Population?

Because

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with Replacementwith Replacement

Each data unit in the

population is allowed to

appear in the sample

more than once

Each data unit in the

population is allowed to

appear in the sample

more than once

Each data unit in the

population is allowed to

appear in the sample

no more than once

Each data unit in the

population is allowed to

appear in the sample

no more than once

Each data unit in the

population

has a known likelihoodof being

included in the sample

Each data unit in the

population

has a known likelihoodof being

included in the sample

Non-Probability SamplingNon-Probability Sampling

Doesnot involverandom selection;

inclusion of an item is

based onconvenience

Doesnot involverandom selection;

inclusion of an item is

based onconvenience

Probability SamplingProbability Sampling

without Replacementwithout Replacement

Techniques8- 6

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MethodsSimple Random

Systematic Random

StratifiedRandom

Cluster

...each item(person) in the population

has an equal chance of being included

items(people) of the population

are arranged in some order.

A random starting point is selected, and

then everykth member of the populationis selected for the sample

a population is

first divided into subgroups, called strata,

and a sample is selected from each strata

a population is

first divided into primary units, and

samples are selected from each unit

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The law firm of Hoya and Associates has five partners.At their weekly partners meeting each reported thenumber of hours they billed their clients last week:

ExampleExamplePartner Hours

Dunn 22

Hardy 26

Kiers 30

Malinowski 26

Tillman 22Iftwo partners are selected randomly

how many different samples are possible?

Iftwo partners are selected randomlyhow many different samples are possible?

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Partner Hours

Dunn 22

Hardy 26

Kiers 30

Malinowski 26

Tillman 22

Objects

5 taken 2 at a time

Using 5C2 Using 5C2

for a Total of 10 Samples!

Iftwo partners are selected randomlyhow many different samples are possible?

Iftwo partners are selected randomlyhow many different samples are possible?

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Partner Hours

Dunn 22

Hardy 26

Kiers 30

Malinowski 26

Tillman 22

Objects

5

5C2 =5C2 =5!

=2!

= 10 Samples= 10 Samples

(5 2!)

Iftwo partners are selected randomlyhow many different samples are possible?Iftwo partners are selected randomly

how many different samples are possible?

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Partners Samples of 2 Mean

1&2

1&3

1&4

1&5

2&3

2&4

2&5

3&4

3&5

4&5

(22+26)/2 =

(22+30)/2 =

(22+26)/2 =

24

2624

(22+22)/2 =

(26+30)/2 =

(26+26)/2 =

(26+22)/2 =

(30+26)/2 =

(30+22)/2 =

(26+22)/2 =

22

28

26

2428

26

24

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Terminology

is the difference betweena sample statistic

and its

corresponding populationparameter

is the difference betweena sample statistic

and its

corresponding populationparameter

is a probability distributionconsisting of

all possible sample meansof a given sample size

selected from a population

is a probability distributionconsisting of

all possible sample meansof a given sample size

selected from a population

Sampling error

Sampling distributionof the sample mean

ExampleExample

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Sample

Mean

Frequency Relative frequency

Probability

Organize the sample meansinto a Sampling Distribution

Organize the sample meansinto a Sampling Distribution

Example continuedExample continuedMean

24

26

24

22

28

26

24

2826

24

22

24

26

28

1

4

3

2

1/10

4/10

3/10

2/10

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Sample Mean Frequency

10

==== 22(1)+ 24(4)+ 26(3) + 28(2)

Example continuedExample continued

22

24

26

28

1

4

3

2

Compute themean of the sample means .Compare it with the population mean

Compute themean of the sample means .Compare it with the population mean

= 25.2= 25.2

X

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Example continuedExample continued

5

2226302622 ++++++++++++++++====

Thepopulation mean is also the same asthesample means25.2 hours!

Thepopulation mean is also the same asthesample means25.2 hours!

Note

Partner Hours

Dunn 22

Hardy 26

Kiers 30

Malinowski 26

Tillman 22

= 25.2= 25.2

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The sampling distribution of the means

ofall possible samples of sizengenerated from the population

will beapproximately normally distributed!

CentralLimit TheoremCentralLimit Theorem

Sampling Distributions:Sampling Distributions:

VarianceVariance 2

/n2

/n

Mean (x)Mean (x)

/ n/ nStandard Deviation

(standard error of the mean)Standard Deviation

(standard error of the mean)X

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samplemeansamplestandarddeviation

sample variancesampleproportion

A point estimate is one value (a single point)that is used to estimate a population parameter

PointEstimatesPointEstimates

More

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PointEstimatesPointEstimates

Population followsthe normal distribution

Population followsthe normal distribution

The sampling distribution

of thesample means also follows

the normal distribution

The sampling distribution

of thesample means also follows

the normal distribution

Probability ofa sample mean

falling within a particular region,

use:

Probability ofa sample mean

falling within a particular region,

use: Z =n

X

Population doesNOTfollowthe normal distribution

Population doesNOTfollowthe normal distribution

If the sample is of at least 30

observations, the sample WILL

follow the normal distribution

If the sample is of at least 30

observations, the sample WILL

follow the normal distribution

Probability ofa sample mean

falling within a particular region,

use:

Probability ofa sample mean

falling within a particular region,

use: Z =n

X

s

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CentralLimit TheoremCentralLimit Theorem

Chart 8 6 Results for Several PopulationsChart 8 6 Results for Several Populations

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Suppose it takes anaverage of 330 minutes

for taxpayers toprepare, copy, andmail an income tax

return form.

Suppose it takes anaverage of 330 minutes

for taxpayers toprepare, copy, andmail an income tax

return form.

Using the Sampling Distribution

of the Sample Mean

Using the Sampling Distribution

of the Sample Mean

= 12.6= 12.6

A consumer watchdogagency selects arandomsample of 40 taxpayersand finds the standarddeviation of the timeneeded is 80 minutes

A consumer watchdogagency selects arandomsample of 40 taxpayersand finds the standarddeviation of the timeneeded is 80 minutes

What is thestandard error of the mean?What is thestandard error of the mean?

Data

/n/nFormulaFormula = 80/= 80/ 40

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What is the likelihood the sample mean

isgreater than 320 minutes?

What is the likelihood the sample mean

isgreater than 320 minutes?

Using the Sampling Distribution

of the Sample Mean

Using the Sampling Distribution

of the Sample Mean

Suppose it takes anaverage of 330 minutes fortaxpayers to prepare, copy, and mail an income taxreturn form. A consumer watchdog agency selects a

random sample of 40 taxpayers and finds thestandard deviation of the time needed is 80 minutes.

Suppose it takes anaverage of 330 minutes fortaxpayers to prepare, copy, and mail an income taxreturn form. A consumer watchdog agency selects a

random sample of 40 taxpayers and finds thestandard deviation of the time needed is 80 minutes.

Data

nswer

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Using the Sampling Distributionof the Sample Mean

Using the Sampling Distributionof the Sample Mean

What is the likelihood the sample meanisgreater than 320 minutes?

What is the likelihood the sample meanisgreater than 320 minutes?

* average of 330 minutes *random sample of 40* standard deviation is 80 minutes

* average of 330 minutes *random sample of 40* standard deviation is 80 minutes

Data

ns

Xz

====FormulaFormula

4080

330320 ==== = 0.79= 0.79

1111

330320

a1

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Using the Sampling Distributionof the Sample Mean

Using the Sampling Distributionof the Sample Mean

What is the likelihood the sample meanisgreater than 320 minutes?

What is the likelihood the sample meanisgreater than 320 minutes?

* average of 330 minutes *random sample of 40* standard deviation is 80 minutes

* average of 330 minutes *random sample of 40* standard deviation is 80 minutes

Data

Look up 0.79in Table

Look up 0.79in Table

2222

a1 =0.2852a1 =0.2852Required Area =

0.2852 + .5 = 0.7852Required Area =

0.2852 + .5 = 0.7852

330320

a1

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Sampling Distribution of

Proportion

Sampling Distribution of

Proportion

The normal distribution(acontinuous distribution)

yields a goodapproximation ofthe binomial distribution

(adiscrete distribution)

for large values ofn.

Use whennp andn(1-p) are both greater than 5!Use whennp andn(1-p) are both greater than 5!

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np=

)1( pnp ====

Mean andVarianceof a

Binomial ProbabilityDistribution

Mean andVarianceof a

Binomial ProbabilityDistribution

2

FormulaFormula

2FormulaFormula

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A multinational company claims that 55% of itsemployees are bilingual. To verify this claim, a

statistician selected a sample of 60 employees of thecompany usingsimple random sampling and

found 48% to be bilingual.

np = 60(.55)= 33

n(1-p) = 60(.45)

= 27

The sample size is bigenough to use the normal

approximation with amean of.55and astandard deviation

of (.55)(.45)/60 = 0.064

The sample size is bigenough to use the normal

approximation with amean of.55and astandard deviation

of (.55)(.45)/60 = 0.064

Sampling Distribution ofProportion

Sampling Distribution ofProportion

Based on this information,what can we say about the companys claim?

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s

Xz

====

1111 Z = (0.48 -0.55) / 0.064Z = -1.09

Look up 1.09 in TableLook up 1.09 in Table22222222

a1 =0.3621a1 =0.3621

Required Area= .5 0.3621 = 0.1379

or 14%

Required Area= .5 0.3621 = 0.1379

or 14%

Sampling Distribution ofProportion

Sampling Distribution ofProportion

continuedcontinued

FormulaFormula

.55.48

a1

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s

Xz

====

1111 Z = (0.48 -0.55) / 0.064Z = -1.09

Look up 1.09 in TableLook up 1.09 in Table22222222

a =0.3621a =0.3621

Required Area= .5 0.3621 = 0.1379

or 14%

Required Area= .5 0.3621 = 0.1379

or 14%

There is

approximately

a 14%chance

that the

companys claim

is true, based onthis sample.

There is

approximately

a 14%chance

that the

companys claim

is true, based onthis sample.

Sampling Distribution of

Proportion

Sampling Distribution of

Proportion

Conclusion

continuedcontinued

FormulaFormula

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Suppose themean selling price of a

litre of gasoline in Canada is$.659.

Further, assume the distribution ispositively

skewed, with astandarddeviationof$0.08.

What is the probability of selecting a

sample of 35 gasoline stations and

finding the sample mean within$.03 ofthe population mean?

Sampling Distribution of

Mean

Sampling Distribution of

Mean

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Sampling Distribution ofMean

Sampling Distribution ofMean

nsz X ====

1 3508.0$

659$.629$. ==== 22.-2====

ns

z X ====2

3508.0$

659$.689$. ==== 2.22====

mean selling price is$.659 SDof $0.08Sample of 35 gasoline stations

Probability ofsample mean within$.03?

mean selling price is$.659 SDof $0.08Sample of 35 gasoline stations

Probability ofsample mean within$.03?

Data

Find thez-scores for.659 +/- .03

Find thez-scores for.659 +/- .03 i.e. 0.629and.689

.629 .689

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We would expect about97%

of the sample means tobe within $0.03 of the

population mean.

We would expect about97%

of the sample means tobe within $0.03 of the

population mean.

a1 = .4868a2 = .4868

Sampling Distribution ofMean

Sampling Distribution ofMean

mean selling price is$.659 SDof $0.08Sample of 35 gasoline stations

Probability ofsample mean within $.03?

mean selling price is$.659 SDof $0.08Sample of 35 gasoline stations

Probability ofsample mean within $.03?

Data

Find areas from tableFind areas from table

Required A = .9736

z ==== -2.221z ==== 2.22

2

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This completes Chapter 8This completes Chapter 8