© 2005 Thomson/South-Western Slide 1 - Governors … · 2009-01-05 · 6 © 2005...

1

1Slide© 2005 Thomson/South-Western

Slides Prepared byJOHN S. LOUCKS

St. Edward’s University


Chapter 8Interval Estimation

� Population Mean: σ Known� Population Mean: σ Unknown� Determining the Sample Size� Population Proportion


A point estimator cannot be expected to provide theexact value of the population parameter.

An interval estimate can be computed by adding andsubtracting a margin of error to the point estimate.

Point Estimate +/− Margin of Error

The purpose of an interval estimate is to provideinformation about how close the point estimate is tothe value of the parameter.

Margin of Error and the Interval Estimate

2


The general form of an interval estimate of apopulation mean is

Margin of Errorx ± Margin of Errorx ±

Margin of Error and the Interval Estimate


Interval Estimation of a Population Mean:σ Known

� In order to develop an interval estimate of a population mean, the margin of error must be computed using either:• the population standard deviation σ , or• the sample standard deviation s

� σ is rarely known exactly, but often a good estimate can be obtained based on historical data or other information.

� We refer to such cases as the σ known case.


There is a 1 − α probability that the value of asample mean will provide a margin of error of or less.

z xα σ/2z xα σ/2

µ

α/2 α/21 - α of allvaluesxx

Samplingdistribution

of xx

xx

z xα σ/2z xα σ/2z xα σ/2z xα σ/2

Interval Estimation of a Population Mean:σ Known

3


µ

α/2 α/21 - α of allvaluesxx


of xx

xx

z xα σ/2z xα σ/2z xα σ/2z xα σ/2[------------------------- -------------------------]

[------------------------- -------------------------][------------------------- -------------------------]

xxxx

xx

intervaldoes notinclude µ interval

includes µinterval

includes µ

Interval Estimate of a Population Mean:σ Known


� Interval Estimate of µ


x zn

± ασ

/2x zn

± ασ

/2

where: is the sample mean1 -α is the confidence coefficientzα/2 is the z value providing an area of

α/2 in the upper tail of the standard normal probability distribution

σ is the population standard deviationn is the sample size

xx



� Adequate Sample Size

In most applications, a sample size of n = 30 isadequate.

If the population distribution is highly skewed orcontains outliers, a sample size of 50 or more isrecommended.

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� Adequate Sample Size (continued)

If the population is believed to be at leastapproximately normal, a sample size of less than 15can be used.

If the population is not normally distributed but isroughly symmetric, a sample size as small as 15 will suffice.


SD

Interval Estimate of Population Mean:σ Known

� Example: Discount SoundsDiscount Sounds has 260 retail outlets

throughout the United States. The firmis evaluating a potential location for anew outlet, based in part, on the meanannual income of the individuals inthe marketing area of the new location.

A sample of size n = 36 was taken;the sample mean income is $31,100. Thepopulation is not believed to be highly skewed. The population standard deviation is estimated to be $4,500,and the confidence coefficient to be used in the interval estimate is .95.


95% of the sample means that can be observedare within + 1.96 of the population mean µ. σxσx

The margin of error is:

ασ ⎛ ⎞= =⎜ ⎟

⎝ ⎠/2

4,5001.96 1,47036

znα

σ ⎛ ⎞= =⎜ ⎟⎝ ⎠

/24,5001.96 1,470

36z

n

Thus, at 95% confidence, the margin of erroris $1,470.

SD


5


Interval estimate of µ is:


SD

We are 95% confident that the interval contains thepopulation mean.

$31,100 + $1,470or

$29,630 to $32,570


Interval Estimation of a Population Mean:σ Unknown

� If an estimate of the population standard deviation σcannot be developed prior to sampling, we use the sample standard deviation s to estimate σ .

� This is the σ unknown case.� In this case, the interval estimate for µ is based on the

t distribution.� (We’ll assume for now that the population is

normally distributed.)


The t distribution is a family of similar probabilitydistributions.

t Distribution

A specific t distribution depends on a parameterknown as the degrees of freedom.

Degrees of freedom refer to the number of independent pieces of information that go into thecomputation of s.

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t Distribution

A t distribution with more degrees of freedom hasless dispersion.

As the number of degrees of freedom increases, thedifference between the t distribution and thestandard normal probability distribution becomessmaller and smaller.


t Distribution

Standardnormal

distribution

t distribution(20 degreesof freedom)

t distribution(10 degreesof freedom)

0z, t


For more than 100 degrees of freedom, the standardnormal z value provides a good approximation tothe t value.

t Distribution

The standard normal z values can be found in theinfinite degrees ( ) row of the t distribution table.∞∞

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t Distribution

Degrees Area in Upper Tailof Freedom .20 .10 .05 .025 .01 .005

. . . . . . .50 .849 1.299 1.676 2.009 2.403 2.67860 .848 1.296 1.671 2.000 2.390 2.66080 .846 1.292 1.664 1.990 2.374 2.639

100 .845 1.290 1.660 1.984 2.364 2.626.842 1.282 1.645 1.960 2.326 2.576∞∞

Standard normalz values


� Interval Estimate

x t sn

± α/2x t sn

± α/2

where: 1 -α = the confidence coefficienttα/2 = the t value providing an area of α/2

in the upper tail of a t distributionwith n - 1 degrees of freedom

s = the sample standard deviation



A reporter for a student newspaper is writing anarticle on the cost of off-campushousing. A sample of 16efficiency apartments within ahalf-mile of campus resulted ina sample mean of $650 per month and a samplestandard deviation of $55.


� Example: Apartment Rents

8


Let us provide a 95% confidence interval estimate of the mean rent permonth for the population of efficiency apartments within ahalf-mile of campus. We willassume this population to be normally distributed.


� Example: Apartment Rents


At 95% confidence, α = .05, and α/2 = .025.

Degrees Area in Upper Tailof Freedom .20 .100 .050 .025 .010 .005

15 .866 1.341 1.753 2.131 2.602 2.94716 .865 1.337 1.746 2.120 2.583 2.92117 .863 1.333 1.740 2.110 2.567 2.89818 .862 1.330 1.734 2.101 2.520 2.87819 .861 1.328 1.729 2.093 2.539 2.861. . . . . . .

In the t distribution table we see that t.025 = 2.131.t.025 is based on n − 1 = 16 − 1 = 15 degrees of freedom.



x t sn

± .025x t sn

± .025

We are 95% confident that the mean rent per monthfor the population of efficiency apartments within ahalf-mile of campus is between $620.70 and $679.30.



55650 2.131 650 29.3016

± = ±55650 2.131 650 29.3016

± = ±

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Summary of Interval Estimation Proceduresfor a Population Mean

Can thepopulation standard

deviation σ be assumed known ?

Use the samplestandard deviation

s to estimate s

Use

Yes No

/2sx tnα± /2

sx tnα±

Use

/2x znα

σ± /2x z

nασ

±

σ KnownCase

σ UnknownCase


Let E = the desired margin of error.

E is the amount added to and subtracted from thepoint estimate to obtain an interval estimate.

Sample Size for an Interval Estimateof a Population Mean



E zn

= ασ

/2E zn

= ασ

/2

n zE

=( )/α σ2

2 2

2n zE

=( )/α σ2

2 2

2

� Margin of Error

� Necessary Sample Size

10


Recall that Discount Sounds is evaluating a potential location for a new retail outlet, based in part, on the mean annual income of the individuals inthe marketing area of the new location.

Suppose that Discount Sounds’ management teamwants an estimate of the population mean such thatthere is a .95 probability that the sampling error is $500or less.

How large a sample size is needed to meet therequired precision?

SD



At 95% confidence, z.025 = 1.96. Recall that σ = 4,500.

znα

σ/2 500=z

nασ

/2 500=

2 2

2

(1.96) (4,500)311.17 312

(500)n = = =

2 2

2

(1.96) (4,500)311.17 312

(500)n = = =


SD

A sample of size 312 is needed to reach a desiredprecision of + $500 at 95% confidence.


The general form of an interval estimate of apopulation proportion is

Margin of Errorp ± Margin of Errorp ±

Interval Estimationof a Population Proportion

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The sampling distribution of plays a key role incomputing the margin of error for this intervalestimate.

pp

The sampling distribution of can be approximatedby a normal distribution whenever np > 5 andn(1 – p) > 5.

pp


α/2 α/2


� Normal Approximation of Sampling Distribution of pp


of pp

(1 )p

p pn

σ−

=(1 )

pp p

nσ

−=

ppp

/2 pzα σ/2 pzα σ/2 pzα σ/2 pzα σ

1 - α of allvaluespp




p z p pn

±−

α /( )

21p z p pn

±−

α /( )

21

where: 1 -α is the confidence coefficientzα/2 is the z value providing an area of

α/2 in the upper tail of the standardnormal probability distributionis the sample proportionpp

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Political Science, Inc. (PSI)specializes in voter polls andsurveys designed to keeppolitical office seekers informedof their position in a race.

Using telephone surveys, PSI interviewers askregistered voters who they would vote for if theelection were held that day.


� Example: Political Science, Inc.


In a current election campaign, PSI has just found that 220registered voters, out of 500contacted, favor a particularcandidate.

PSI wants to develop a 95% confidence interval estimate for the proportion of the population ofregistered voters that favor the candidate.


� Example: Political Science, Inc.


p z p pn

±−

α/( )

21p z p pn

±−

α/( )

21

where: n = 500, = 220/500 = .44, zα/2 = 1.96pp


PSI is 95% confident that the proportion of all votersthat favor the candidate is between .3965 and .4835.

.44(1 .44).44 1.96500

−±

.44(1 .44).44 1.96500

−± = .44 + .0435

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Solving for the necessary sample size, we get

� Margin of Error

Sample Size for an Interval Estimateof a Population Proportion

/2(1 )p pE z

nα−

= /2(1 )p pE z

nα−

=

2/2

2

( ) (1 )z p pnE

α −=

2/2

2

( ) (1 )z p pnE

α −=

However, will not be known until after we have selected the sample. We will use the planning valuep* for .

pp

pp



The planning value p* can be chosen by:1. Using the sample proportion from a previous

sample of the same or similar units, or2. Selecting a preliminary sample and using the

sample proportion from this sample.

2 * */2

2

( ) (1 )z p pnE

α −=

2 * */2

2

( ) (1 )z p pnE

α −=

� Necessary Sample Size


Suppose that PSI would like a .99 probabilitythat the sample proportion is within + .03 of the population proportion.

How large a sample size is needed to meet the required precision? (A previous sample of similar units yielded .44 for the sample proportion.)


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At 99% confidence, z.005 = 2.576. Recall that = .44.pp

A sample of size 1817 is needed to reach a desiredprecision of + .03 at 99% confidence.

2 2/2

2 2

( ) (1 ) (2.576) (.44)(.56) 1817(.03)

z p pn

Eα −

= = ≅2 2

/22 2

( ) (1 ) (2.576) (.44)(.56) 1817(.03)

z p pn

Eα −

= = ≅


/ 2(1 ) .03p pz

nα−

=/ 2(1 ) .03p pz

nα−

=


Note: We used .44 as the best estimate of p in thepreceding expression. If no information is availableabout p, then .5 is often assumed because it providesthe highest possible sample size. If we had usedp = .5, the recommended n would have been 1843.



End of Chapter 8

© 2005 Thomson/South-Western Slide 1 - Governors … · 2009-01-05 · 6 © 2005...

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