Simple Random Sampling Point Estimation Introduction to...

Chapter 7Sampling and Sampling Distributions

xxSampling Distribution ofIntroduction to Sampling DistributionsPoint EstimationSimple Random Sampling

Example: St. Andrew’s

St. Andrew’s College receives900 applications annually fromprospective students. Theapplication form contains a variety of informationincluding the individual’sscholastic aptitude test (SAT) score and whether or notthe individual desires on-campus housing.


The director of admissionswould like to know thefollowing information:• the average SAT score for

the 900 applicants, and• the proportion of

applicants that want to live on campus.


We will now look at threealternatives for obtaining thedesired information.� Conducting a census of the

entire 900 applicants� Selecting a sample of 30

applicants, using a random number table� Selecting a sample of 30 applicants, using Excel

Conducting a Census

� If the relevant data for the entire 900 applicants were in the college’s database, the population parameters of interest could be calculated using the formulas presented in Chapter 3.

� We will assume for the moment that conducting a census is practical in this example.

990900

ixμ = =∑ 990900

ixμ = =∑

2( )80

900ix μσ−

= =∑2( )

80900

ix μσ−

= =∑

Conducting a Census

648 .72900

p = =648 .72900

p = =

� Population Mean SAT Score

� Population Standard Deviation for SAT Score

� Population Proportion Wanting On-Campus Housing

� as Point Estimator of μxx

� as Point Estimator of ppp

29,910 99730 30

ixx = = =∑ 29,910 99730 30

ixx = = =∑

2( ) 163,996 75.229 29ix xs−

= = =∑2( ) 163,996 75.2

29 29ix xs−

= = =∑

20 30 .68p = =20 30 .68p = =

Point Estimation

Note: Different random numbers would haveidentified a different sample which would haveresulted in different point estimates.

� s as Point Estimator of σ

PopulationParameter

PointEstimator

PointEstimate

ParameterValue

μ = Population meanSAT score

990 997

σ = Population std.deviation for SAT score

80 s = Sample std.deviation forSAT score

75.2

p = Population pro-portion wantingcampus housing

.72 .68

Summary of Point EstimatesObtained from a Simple Random Sample

= Sample meanSAT score

xx

= Sample pro-portion wantingcampus housing

pp

� Process of Statistical Inference

The value of is used tomake inferences about

the value of μ.

xx The sample data provide a value forthe sample mean .xx

A simple random sampleof n elements is selected

from the population.

Population with mean

μ = ?

Sampling Distribution of xx

The sampling distribution of is the probabilitydistribution of all possible values of the sample mean .

xx

xx


where: μ = the population mean

E( ) = μxx

xxExpected Value of


Finite Population Infinite Population

σ σx nN nN

=−−

( )1

σ σx nN nN

=−−

( )1

σ σx n=σ σx n=

• is referred to as the standard error of themean.σ xσ x

• A finite population is treated as beinginfinite if n/N < .05.

• is the finite correction factor.( ) / ( )N n N− −1( ) / ( )N n N− −1

xxStandard Deviation of

Form of the Sampling Distribution of xx

If we use a large (n > 30) simple random sample, thecentral limit theorem enables us to conclude that thesampling distribution of can be approximated bya normal distribution.

xx

When the simple random sample is small (n < 30),the sampling distribution of can be considerednormal only if we assume the population has anormal distribution.

xx

80 14.630x n

σσ = = =80 14.630x n

σσ = = =

( ) 990E x =( ) 990E x =xx

Sampling Distribution of for SAT Scoresxx

SamplingDistribution

of xx

With a mean SAT score of 990 and a standard deviation of80, what is the probability that a simple random sampleof 30 applicants will provide an estimate of thepopulation mean SAT score that is within +/−10 ofthe actual population mean μ ?

In other words, what is the probability that will bebetween 980 and 1000?

xx


Step 1: Calculate the z-value at the upper endpoint ofthe interval.

z = (1000 - 990)/14.6= .68

.2517

Step 2: Find the area under the curve between the meanand the upper endpoint.



Probabilities forthe Standard Normal Distribution

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09. . . . . . . . . . ..5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389. . . . . . . . . . .

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09. . . . . . . . . . ..5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389. . . . . . . . . . .


xx990


of xx14.6xσ = 14.6xσ =

1000

Area = .2517

Step 3: Calculate the z-value at the lower endpoint ofthe interval.

Step 4: Find the area under the curve between the m ean and the lower endpoint.

z = (980 - 990)/14.6= - .68

= .2517



xx990


of xx14.6xσ = 14.6xσ =

980

Area = .2517


xx980 990

Area = .2517


of xx14.6xσ = 14.6xσ =

1000

Area = .2517


Step 5: Calculate the area under the curve betweenthe lower and upper endpoints of the interval.

P(-.68 < z < .68) == .2517 + .2517= .5034

The probability that the sample mean SAT score willbe between 980 and 1000 is:

P(980 < < 1000) = .5034xx

xx1000980 990


Area = .5034


of xx14.6xσ = 14.6xσ =

Relationship Between the Sample Sizeand the Sampling Distribution of xx

Suppose we select a simple random sample of 100applicants instead of the 30 originally considered.

E( ) = μ regardless of the sample size. In ourexample, E( ) remains at 990.

xxxx

Whenever the sample size is increased, the standarderror of the mean is decreased. With the increasein the sample size to n = 100, the standard error of themean is decreased to:

xσ xσ

80 8.0100x n

σσ = = =80 8.0100x n

σσ = = =


( ) 990E x =( ) 990E x =xx

14.6xσ = 14.6xσ =With n = 30,

8xσ = 8xσ =With n = 100,

Recall that when n = 30, P(980 < < 1000) = .5034.xx


We follow the same steps to solve for P(980 < < 1000)when n = 100 as we showed earlier when n = 30.

xx

Now, with n = 100, P(980 < < 1000) = .7888.xxBecause the sampling distribution with n = 100 has asmaller standard error, the values of have lessvariability and tend to be closer to the populationmean than the values of with n = 30.

xx

xx


xx1000980 990

Area = .7888


of xx8xσ = 8xσ =

Chapter 7 Sampling and Sampling Distributions

Other Sampling Methods

ppSampling Distribution of

A simple random sampleof n elements is selected

from the population.

Population with proportion

p = ?

� Making Inferences about a Population Proportion

The sample data provide a value for thesample proportion .pp

The value of is usedto make inferences

about the value of p.

pp

Sampling Distribution of pp

E p p( ) =E p p( ) =


where:p = the population proportion

The sampling distribution of is the probabilitydistribution of all possible values of the sampleproportion .pp

pp

ppExpected Value of

σ pp p

nN nN

=− −

−( )1

1σ p

p pn

N nN

=− −

−( )1

1σ p

p pn

=−( )1σ p

p pn

=−( )1

is referred to as the standard error of theproportion.σ pσ p


Finite Population Infinite Population

ppStandard Deviation of

• A finite population is treated as beinginfinite if n/N < .05.

Recall that 72% of theprospective students applyingto St. Andrew’s College desireon-campus housing.

� Example: St. Andrew’s College


What is the probability thata simple random sample of 30 applicants will providean estimate of the population proportion of applicantdesiring on-campus housing that is within plus orminus .05 of the actual population proportion?

σ −= =p.72(1 .72) .082

30σ −= =p

.72(1 .72) .08230

( ) .72E p =( ) .72E p =pp


of pp


Step 1: Calculate the z-value at the upper endpoint ofthe interval.

z = (.77 - .72)/.082 = .61

.2291

Step 2: Find the area under the curve between the m ean and upper endpoint.


Probabilities forthe Standard Normal Distribution

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09. . . . . . . . . . ..5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389. . . . . . . . . . .

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09. . . . . . . . . . ..5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2517 .2549.7 .2580 .2611 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389. . . . . . . . . . .


.77.72

Area = .2291

pp


of pp

.082pσ = .082pσ =


Step 3: Calculate the z-value at the lower endpoint ofthe interval.

Step 4: Find the area under the curve between the m ean and the lower endpoint.

z = (.67 - .72)/.082 = - .61

.2291


.67 .72

Area = .2291

pp


of pp

.082pσ = .082pσ =


P(.67 < < .77) = .4582pp

Step 5: Calculate the area under the curve betweenthe lower and upper endpoints of the interval.

P(-.61 < z < .61) == .2291 + .2291= .4582

The probability that the sample proportion of applicantswanting on-campus housing will be within +/-.05 of theactual population proportion :


.77.67 .72

Area = .4582

pp


of pp

.082pσ = .082pσ =


Other Sampling Methods

� Stratified Random Sampling� Cluster Sampling� Systematic Sampling� Convenience Sampling� Judgment Sampling

The population is first divided into groups ofelements called strata.

Stratified Random Sampling

Each element in the population belongs to one andonly one stratum.

Best results are obtained when the elements withineach stratum are as much alike as possible(i.e. a homogeneous group).

Stratified Random Sampling

A simple random sample is taken from each stratum.

Formulas are available for combining the stratumsample results into one population parameterestimate.

Advantage: If strata are homogeneous, this methodis as “precise” as simple random sampling but witha smaller total sample size.

Example: The basis for forming the strata might bedepartment, location, age, industry type, and so on.

Cluster Sampling

The population is first divided into separate groupsof elements called clusters.

Ideally, each cluster is a representative small-scaleversion of the population (i.e. heterogeneous group).

A simple random sample of the clusters is then taken.

All elements within each sampled (chosen) clusterform the sample.

Cluster Sampling

Advantage: The close proximity of elements can becost effective (i.e. many sample observations can beobtained in a short time).

Disadvantage: This method generally requires alarger total sample size than simple or stratifiedrandom sampling.

Example: A primary application is area sampling,where clusters are city blocks or other well-definedareas.

Systematic Sampling

If a sample size of n is desired from a populationcontaining N elements, we might sample oneelement for every n/N elements in the population.

We randomly select one of the first n/N elementsfrom the population list.

We then select every n/Nth element that follows inthe population list.

Systematic Sampling

This method has the properties of a simple randomsample, especially if the list of the populationelements is a random ordering.

Advantage: The sample usually will be easier toidentify than it would be if simple random samplingwere used.

Example: Selecting every 100th listing in a telephonebook after the first randomly selected listing

Convenience Sampling

It is a nonprobability sampling technique. Items areincluded in the sample without known probabilitiesof being selected.

Example: A professor conducting research might usestudent volunteers to constitute a sample.

The sample is identified primarily by convenience.

Advantage: Sample selection and data collection arerelatively easy.

Disadvantage: It is impossible to determine howrepresentative of the population the sample is.

Convenience Sampling

Judgment Sampling

The person most knowledgeable on the subject of thestudy selects elements of the population that he orshe feels are most representative of the population.

It is a nonprobability sampling technique.

Example: A reporter might sample three or foursenators, judging them as reflecting the generalopinion of the senate.

Judgment Sampling

Advantage: It is a relatively easy way of selecting asample.

Disadvantage: The quality of the sample resultsdepends on the judgment of the person selecting thesample.

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