Sampling Distribution[1]

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Sampling Methods and

Sampling Distr ibutionsChapter


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2EXPLAIN WHY SAMPLES ARE USED.

DEFINE AND CONSTRUCT A SAMPLINGDISTRIBUTION OF SAMPLE MEANS.

EXPLAIN THE CENTRAL LIMIT THEOREM

CALCULATE CONFIDENCE INTERVALSFOR MEANS AND PROPORTIONS.

DETERMINE HOW LARGE A SAMPLESHOULD BE FOR BOTH MEANS ANDPROPORTIONS.

GOALS


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3 The destructive nature of certain tests.

The physical impossibility of checking allitems in the population.

The cost of studying all the items in apopulation is often prohibitive.

The adequacy of sample results.

To contact the whole population wouldoften be time-consuming.

WHY SAMPLE THE POPULATION?


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4What is a Probabi l i ty Sample?

A sample selected in such a way that eachitem or person in the population beingstudied has a known (nonzero) likelihood of

being included in the sample. Simple Random Sample:A sample

formulated so that each item or person in

the population has the same chance ofbeing included.

PROBABILITY SAMPLING


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Given a list of elements, select a randomsubset.

Tools

Uniform Random Number Generator Sort Function

How?

SIMPLE RANDOM SAMPLING VIA EXCEL


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6

Systemat ic Random Sampl ing:The items orindividuals of the population are arrangedin some way-alphabetically or by someother method. A random starting point is

selected, and then every kth member of thepopulation is selected for the sample.

Strat i f ied Random Sampl ing :A population

is first divided into subgroups, calledstrata, and a sample is selected from eachstratum.

PROBABILITY SAMPLING (continued)


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Sampl ing Error :The difference between asample statistic and its correspondingparameter.

For example

PROBABILITY SAMPLING (continued)


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A probability distribution consisting of a listof all possible sample means of a givensample size selected from a population, andthe probability of occurrence associated

with each sample mean.

EXAMPLE :A law firm has five partners. Attheir weekly partners meeting each

reported the number of hours they chargedclients for their professional services lastweek. The results are given on the next

slide.

SAMPLING DISTRIBUTION OF

THE SAMPLE MEANS


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9EXAMPLE (continued)

Two partners are randomly selected. Howmany different samples are possible?


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10 This is the combination of 5 objects taken 2

at a time. That is, 5C2 = (5!)/[(2!)(3!)] = 10. List the possible samples of size 2 and

compute the mean.

EXAMPLE (continued)


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11Organize the sample means into a sampling

distribution. The sampling distribution isshown below.

EXAMPLE (continued)


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12Compute the mean of the sample means

and compare it with the population mean. The population mean,

m = (22 + 26 + 30 + 26 + 22)/5 = 25.2. The mean of the sample means = [(22)(1) +

(24)(4) + (26)(3) + (28)(2)]/10 = 25.2.

Observe that the mean o f the sample means

is equal to the popu lat ion mean.

EXAMPLE (cont inued)


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13 For a population with a mean m and a

variance s2, the sampling distribution of themeans of all possible samples of size ngenerated from the population will be

approximately normally distributed - withthe mean of the sampling distribution equalto m and the variance equal to s2/n-assuming that the sample size issufficiently large.

CENTRAL LIMIT THEOREM


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14POINT ESTIMATESOne value (called a po in t) that is used to

estimate a population parameter. Examples of point estimates are the samplemean, the sample standard deviat ion, the

sample var iance, the sample propo rt ion etc. EXAMPLE: The number of defective items

produced by a machine was recorded for fiverandomly selected hours during a 40-hourwork week. The observed number ofdefectives were 12, 4, 7, 14, and 10. So thesample mean is 9.4. Thus a point estimate for

the weekly mean number of defectives is 9.4


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15INTERVAL ESTIMATESAn In terval Estim atestates the range within

which a population parameter probably lies. The interval within which a population

parameter is expected to occur is called a

conf idence interval. The two confidence intervals that are used

extensively are the 95% and the 99%.

A 95%confidence interval means that about95% of the similarly constructed intervalswill contain the parameterbeing estimated.


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16INTERVAL ESTIMATES (con tinued)Another interpretation of the 95%

confidence interval is that 95% of thesample means for a specified sample sizewill lie within 1.96 standard deviations of

the hypothesized population mean. For the 99% confidence interval, 99% of the

sample means for a specified sample size

will lie within 2.58 standard deviations ofthe hypothesized population mean.


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17

-2.58 -1.96 1.96 2.58

95%

99%

0

STANDARD ERROR OF THE SAMPLE


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18STANDARD ERROR OF THE SAMPLE

MEANS

This is the standard deviation of thesampling distribution of the sample means.

The standard errorof the sample means iscomputed by:

is the symbol for the standard error ofthe sample means.

sis the standard deviation of the population.n is the size of the sample.

ss

xn

sx



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Ifsis not knownand n= 30 ormore(considered a large sample), the standarddeviation of the sample, designated by s, isused to approximate the population

standard deviation, s. The formula for thestandard error then becomes:

What happens as n gets larger?

s

s

n

x


MEANS (continued)

95% AND THE 99% CONFIDENCE


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2095% AND THE 99% CONFIDENCE

INTERVALS (CI) FOR m

The 95% and the 99% confidence intervalsform are constructed as follows when n30.

95% CI for the population meanmis givenby

99% CI formis given by

Xs

n196.

Xs

n

258.

CONSTRUCTING A GENERAL


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In general, a confidence interval for themean is computed by:

The Z value is obtained from the standard

no rmal table in Append ix D (look-upconfidence/2).

CONSTRUCTING A GENERAL

CONFIDENCE INTERVALS (CI) FORm

X Zs

n


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22 The Dean of Students at Penta Tech wants to

estimate the mean number of hours workedper week by students. A sample of 49students showed a mean of 24 hours with astandard deviation of 4 hours.

What is the point estimate of the mean numberof hours worked per week by students?

The point estimate is 24 hours (sample mean).

What is the 95% confidence interval for theaverage number of hours worked per week bythe students?

EXAMPLE


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23Using formula, we have 24 1.96(4/7) or we

have 22.88 to 25.12.What are the 95% conf idence l imi ts?

The endpoints of the confidence interval

are the confidence limits. The lowerconf idence l im i tis 22.88 and the upperconf idence l im i tis 25.12.

What degree of conf idenceis being used? The degree of confidence (level of

confidence) is 0.95

EXAMPLE (continued)


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24 Interpret the findings.

If we had time to select 100 samples of size 49from the population of the number of hours worked

per week by students at Penta Tech and compute

the sample means and 95% confidence intervals,

the population mean of the number of hours

worked by the students per week would be found in

about 95 out of the 100 confidence intervals.

Either a confidence interval contains the populationmean or it does not. About 5 out of the 100

confidence intervals would not contain the

population mean.

EXAMPLE(cont inued)

CONFIDENCE INTERVAL FOR A


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25CONFIDENCE INTERVAL FOR APOPULATION PROPORTION

The confidence interval for a populationproportion:

where is the standard error of the

proportion:

p z p s

sp

p p

n

( )1

sp

Th fid i t l i t t d b


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26 The confidence interval is constructed by:

where:is the sample proportion.

z is the zvalue for the degree of

confidence selected.n is the sample size.

p z p pn

( )1

p


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27EXAMPLE Chris Cooper, a financial planner, is studying

the retirement plans of young executives. Asample of 500 young executives who ownedtheir own home revealed that 175 planned tosell their homes and retire to Arizona. Develop

a 98% confidence interval for the proportion ofexecutives that plan to sell and move toArizona.

Here n= 500, = 175/500 = 0.35, and z=2.33 the 98% CI is 0.35 2.33 or

0.35 0.0497. Interpret?

p( . )( . )035 065

500

FINITE POPULATION


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28FINITE-POPULATIONCORRECTION FACTOR

A population that has a fixed upper boundis said to be f in i te.

For a f in i te popu lat ion, where the totalnumber of objects is Nand the size of thesample is n, the following adjustment ismade to the standard errors of the samplemeans and the proportion.

Standard error o f the sample means :

s

s

x

n

N n

n

1

FINITE POPULATION


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Standard error of the sample proport ions :

Note: If n /N < 0.05, the fini te-popu lation

co rrect ion factor can be igno red.

sp

p p

n

N n

N

( )1

1

FINITE-POPULATION

CORRECTION FACTOR (con tinued)

EXAMPLE


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30 The Dean of Students at Penta Tech wants to

estimate the mean number of hours workedper week by students. A sample of 49students showed a mean of 24 hours with astandard deviation of 4 hours. Construct a

95% confidence interval for the mean numberof hours worked per week by the students ifthere are only 500 students on campus.

Now n /N= 49/500 = 0.098 > 0.05, so we have touse the finite population correction factor.

= [22.9352,25.1065]

EXAMPLE

24 1964

49

500 49

500 1

.


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31SELECTING A SAMPLE SIZE There are 3 factors that determine the size

of a sample, none of which has any d irectrelat ionship to the size of the popu lat ion .

They are:

1. The degree of confidence selected.2. The maximum allowable error.

3. The variation of the population.


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32SAMPLE SIZE FOR THE MEANA convenient computational formula for

determining n is:

where:

E is the allowable error.

z is the z score associated with the degreeof confidence selected.

s is the sample deviation of the pilotsurvey.

n Z S

E

2


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33A consumer group would like to estimate

the meanmonthly electric bill for a singlefamily house in July. Based on similarstudies the standard deviation is estimated

to be $20.00. A 99% level of confidence isdesired, with an accuracy of $5.00. Howlarge a sample is required?

n= [(2.58)(20)/5]

2

= 106.5024 107.

EXAMPLE

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34SAMPLE SIZE FOR PROPORTIONS The formula for determining the sample

size in the case of a proportion is:

is the estimated proportion, based on pastexperience or a pilot survey.

z is the zvalue associated with the degree ofconfidence selected.

E is the maximum allowable error the researcher will

tolerate.

p

n p p Z

E

( )12

35EXAMPLE


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35 The American Kennel Club wanted to

estimate the propor t ionof children thathave a dog as a pet. If the club wanted theestimate to be within 3% of the population

proportion, how many children would theyneed to contact? Assume a 95% level ofconfidence and that the Club estimated that30% of the children have a dog as a pet.

n= (0.30)(0.70)(1.96/0.03)2 = 896.3733 897.

EXAMPLE

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