chapter 8 interval estimation

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Chapter 8

Interval Estimation

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Chapter Outline

Population Mean: Known Population Mean: Unknown Population Proportion

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Introduction

The sampling distribution introduced last chapter connects sample statistics to population parameters.

In reality, we probably don’t know any of the population parameters. However, a study on sampling distributions can provide reasonable references to the population parameters.

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Margin of Error and the Interval Estimate

A point estimator cannot be expected to provide the exact value of the population parameter. For instance, the probability of any particular sample mean equal population mean is zero, i.e. p( =

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

The purpose of an interval estimate is to provide a reasonable value range of the population parameters.

x

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Margin of Error and the Interval Estimate

The general form of an interval estimates of a population mean is

Error ofMargin x

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Interval Estimate of A Population Mean: Known

In the first scenario, we assume to be known. Although is rarely known in reality, a good estimate can be obtained based on historical data or other information.

Let’s use the example of Checking Accounts from last chapter as an illustration. Here, we assume that the population standard deviation is known (=66). Our goal is to come up with an interval estimate of population mean based on the sample mean =280.x

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Summary of Point Estimates of A Simple Random Sample of 121 Checking Accounts

PopulationPopulationParameterParameter

PointPointEstimatorEstimator

PointPointEstimateEstimate

ParameterParameterValueValue

= Population mean= Population mean account balance account balance

$310$310 $306$306

= Population std.= Population std. deviation for deviation for account balance account balance

$66$66 s s = Sample std.= Sample std. deviation fordeviation for account balance account balance

$61$61

pp = Population pro- = Population proportion of accountportion of account balance no less thanbalance no less than $500 $500

.3.3 .27.27

= Sample mean= Sample mean account balance account balance

xx

= Sample pro-= Sample proportion of accountportion of account balance no less thanbalance no less than $500 $500

pp

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The sample mean distribution of 121 checking account balances can be approximated by a normal distribution with EE( ) = ( ) = Let’s first figure out the values of that Let’s first figure out the values of that provide the middle area about provide the middle area about of 95%. of 95%.

xx

x xE

95%

a b

6121

66x

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Example: Checking Accounts Given the middle area of 95%, we can find z values first and then convert z values to the corresponding values of .

x

x

95%

a b

95%

-z0.025 z0.025

z0

2.5%

2.5%

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Example: Checking Accounts Convert z values to the corresponding values of . x

x

bz

025.0 xzb 025.0

x

az

025.0 xza 025.0

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Example: Checking Accounts We set the margin of error as . So, the interval estimate of population mean is .

x

95%

a b

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

xz 025.0 xz 025.0

xz 025.0

xzx 025.0

xzx 025.0

[--------- -----------][--------- -----------]xzx 025.0

[--------- -----------][--------- -----------]xzx 025.0

As long as falls between a and b, the interval will include the population mean .

xxzx 025.0

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Example: Checking AccountsThe rationale behind the interval estimate – For any particular sample mean , we cannot compare it with the population mean since is unknown. But, what we are certain is that as long as falls between a and b. The interval will include the true value of . In the example, = 306. So, the interval estimate of population account balance is . Because z0.025=1.96 and , the interval estimate is calculated as 3061.96·6 = 306 11.76

or $294.24 to $317.76

We are 95% confident that will fall between a and b. So, the chance is 95% that the true value of is no less than $294.24 and no more than $317.76. On the other hand, there is a 5% chance that we make a mistake and the above interval estimate doesn’t include .

xzx 025.0

xz 025.0306

x

x

6xx

x

Margin of Error

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x zn

/2x zn

/2

Interval Estimate ofInterval Estimate of

where: is the sample meanwhere: is the sample mean 1 -1 - is the confidence level is the confidence level zz/2 /2 is the is the zz value providing an area of value providing an area of /2 in the upper tail of the standard /2 in the upper tail of the standard

normal probability distributionnormal probability distribution is the population standard deviationis the population standard deviation nn is the sample size is the sample size

xx

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Values of Values of zz/2/2 for the Most Commonly Used for the Most Commonly Used Confidence LevelsConfidence Levels

90% .10 .05 1- 90% .10 .05 1- .9500 .9500 1.645 1.645 95% .05 .025 1- 95% .05 .025 1- .9750 .9750 1.960 1.960 99% .01 .005 1- 99% .01 .005 1- .9950 .9950 2.576 2.576

Confidence Area to the Confidence Area to the Level Level /2 left of z/2 left of z/2 /2

zz/2/2

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Example: Checking AccountsExample: Checking Accounts

90% 9.87 296.13 to 90% 9.87 296.13 to 315.87 315.87

Confidence MarginConfidence Margin Level of Error Interval EstimateLevel of Error Interval Estimate

95% 11.76 294.24 to 95% 11.76 294.24 to 317.76 317.76 99% 15.46 290.54 to 99% 15.46 290.54 to 321.46321.46

The higher the confidence level, the wider the The higher the confidence level, the wider the

Interval estimate.Interval estimate.

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Interval Estimate of A Population Mean: Unknown

When is unknown, we will have to use the sample is unknown, we will have to use the sample standard deviation standard deviation s s to estimate to estimate . .

In this case, the interval estimate for is based on the is based on the t t distributiondistribution. (. (See Table 2 of Appendix B in the textbookSee Table 2 of Appendix B in the textbook))• A specific A specific t t distribution depends on a parameter known as the distribution depends on a parameter known as the

degrees of freedom.degrees of freedom.

• Degrees of freedom refer to the number of independent pieces of Degrees of freedom refer to the number of independent pieces of information that go into the computation of information that go into the computation of ss. .

• As the degrees of freedom increases, As the degrees of freedom increases, t t distribution is approaching distribution is approaching closer to the Standard Normal Distribution.closer to the Standard Normal Distribution.

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t Distribution

StandardStandardnormalnormal

distributiondistribution

tt distributiondistribution(20 degrees(20 degreesof freedom)of freedom)

tt distributiondistribution(10 degrees(10 degrees

of of freedom)freedom)

00zz, , tt

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t Distribution For more than 100 degrees of freedom, the standard

normal z value provides a good approximation to the t value.

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

Degrees Area in Upper Tail

of Freedom .20 .10 .05 .025 .01 .005

. . . . . . .

50 .849 1.299 1.676 2.009 2.403 2.678

60 .848 1.296 1.671 2.000 2.390 2.660

80 .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 Standard normalnormalzz values values

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Interval EstimateInterval Estimate

where: 1 -where: 1 - = the confidence level = the confidence level

tt/2 /2 == the the tt value providing an area of value providing an area of /2/2 in the upper tail of a in the upper tail of a t t distribution distribution with with nn - 1 degrees of freedom - 1 degrees of freedom ss = the sample standard deviation = the sample standard deviation

n n = sample size= sample size

n

stx 2/

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Example: Consumer AgeExample: Consumer Age

. The makers of a soft drink want to identify the average age of its consumers. A sample of 20 consumers was taken. The average age in the sample was 21 years with a standard deviation of 4 years.

Construct a 95% confidence interval for the true average age of the consumers.

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

In the In the tt distribution table we see that distribution table we see that tt.025.025 = 2.093. = 2.093.tt.025.025 is based on is based on nn - 1 = 20 - 1 = 19 degrees of freedom. - 1 = 20 - 1 = 19 degrees of freedom.

Degrees Area in Upper Tail

of Freedom .20 .100 .050 .025 .010 .005

15 .866 1.341 1.753 2.131 2.602 2.947

16 .865 1.337 1.746 2.120 2.583 2.921

17 .863 1.333 1.740 2.110 2.567 2.898

18 .862 1.330 1.734 2.101 2.520 2.878

19 .861 1.328 1.729 2.093 2.539 2.861

. . . . . . .

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x tsn

.025x tsn

.025

87.12120

4093.221

Margin of Error

We are 95% confident that the average age of the We are 95% confident that the average age of the soft drink consumers is between 19.13 and 22.87.soft drink consumers is between 19.13 and 22.87.

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Summary of Interval Estimation Summary of Interval Estimation ProceduresProcedures

for a Population Meanfor a Population Mean

Is the Is the population standardpopulation standard

deviation deviation known ?known ?

UseUse

YesYes NoNo

/ 2

sx t

n / 2

sx t

nUseUse/ 2x z

n

/ 2x zn

KnownKnownCaseCase

UnknownUnknownCaseCase

Use the sampleUse the samplestandard deviationstandard deviation

ss to estimate to estimate ss

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Interval Estimate of A Population Proportion

The general form of an interval estimate of a population proportion is

Error ofMargin p

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Just as the sampling distribution of is key in estimating population mean, the sampling distribution of is crucial in estimating population proportion.

The sampling distribution of can be approximated by a normal distribution whenever np 5 and n(1-p) 5.

x

p

p

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/2/2/2/2

Normal Approximation of Sampling Distribution Normal Approximation of Sampling Distribution of of

SamplingSamplingdistributiondistribution of of

SamplingSamplingdistributiondistribution of of

pp

(1 )p

p p

n

(1 )p

p p

n

pppppp

/ 2 pz / 2 pz / 2 pz / 2 pz

1 - 1 -

pp

/2/2/2/2

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Interval Estimate of Interval Estimate of pp

n

ppzp

12/

where: 1 -where: 1 - is the confidence level is the confidence level

zz/2 /2 is the is the zz value providing an area of value providing an area of

/2 in the upper tail of the standard/2 in the upper tail of the standard

normal probability distributionnormal probability distribution

is the sample proportionis the sample proportionpp

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Example: Checking Accounts Refer to our previous example of Checking Accounts. Out of the simple random sample of 121 accounts, the sample proportion of account

balance no less than $500 is .27. Develop a 95% confidence interval estimate of the population proportion.

n

ppzp

12/

where: where: nn = 121, = .27, = 121, = .27, zz/2 /2 = = 1.961.96p

04.27.

121

27.127.96.127.

We are 95% confident that the proportion of We are 95% confident that the proportion of allall checking accounts checking accounts with a balance no less than $500 is between .23 and .31, which with a balance no less than $500 is between .23 and .31, which correctly includes the population proportion .30.correctly includes the population proportion .30.

Margin of Error

chapter 8 interval estimation

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