10. Parameter Estimation

7/28/2019 10. Parameter Estimation

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PARAMETER

ESTIMATION


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Parameters Populations are described by their

probability distributions and/or parameters.

For quantitative populations, the locationand shape are described by mand s.

Binomial populations are determined bya single parameter,p.

If the values of parameters are unknown,we make inferences about them usingsample information.


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Statistical inference is the process bywhich we acquire information aboutpopulations from samples.

There are two types of inference:

Estimation

Hypotheses testing


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Types of Inference

Estimation

Estimating or predicting the value of theparameter

What is (are) the most likely values of

m orp? Hypothesis Testing

Deciding about the value of a parameter

based on some preconceived idea. Did the sample come from a population

with m = 5 orp= .2?


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Types of Inference

Examples:

A consumer wants to estimate the averageprice of similar homes in her city beforeputting her home on the market.

Estimation: Estimate m, the average home price.

Hypothesis test: Is the new average resistance, mN greater thanthe old average resistance, mO?

A manufacturer wants to know if a new typeof steel is more resistant to high temperatures

than an old type was.


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Types of Inference

Whether you are estimating parameters ortesting hypotheses, statistical methods

are important because they provide: Methods for making the inference

A numerical measure of the goodness or

reliability of the inference


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Concepts of Estimation

The objective of estimation is to determinethe value of a population parameter on thebasis of a sample statistic.

There are two types of estimators: Point Estimator Interval estimator


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Definitions

An estimatoris a rule, usually a formula, thattells you how to calculate the estimate based onthe sample.

Point estimation:A single number iscalculated to estimate the parameter.

Interval estimation/Confidence Interval:Two numbers are calculated to create an

interval within which the parameter isexpected to lie. It is constructed so that, witha chosen degree of confidence, the trueunknown parameter will be captured inside

the interval.


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Point Estimator

A point estimator draws inference about apopulation by estimating the value of anunknown parameter using a single value

or point.


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Point Estimator of Population Mean

A sample of weights of 34 male freshman students was obtained.

185 161 174 175 202 178 202 139 177170 151 176 197 214 283 184 189 168

188 170 207 180 167 177 166 231 176

184 179 155 148 180 194 176If one wanted to estimate the true mean of all male freshman students, you might

use the sample mean as a point estimate for the true mean.

sample mean x 182.44= =

n

xxi=

An point estimate of population mean, m, is the

sample mean


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Point Estimation of Population

Proportion

A sample of 200 students at a large university isselected to estimate the proportion of studentsthat wear contact lens. In this sample 47 wearcontact lens.

An point estimate of population mean, p, is thesample proportion where x is the numberof successes in the sample.

nxp /=

235.200/47 ==p


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Population distribution

Point Estimator

Parameter

?

Sampling distribution

A point estimator draws inference about apopulation by estimating the value of anunknown parameter using a single value

or point.

Point estimator


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Properties of

Point Estimators

Since an estimator is calculated from sample values, itvaries from sample to sample according to itssampling distribution.

An estimatoris unbiased if the mean of its sampling

distribution equals the parameter of interest. It doesnot systematically overestimate or underestimate thetarget parameter.

Both sample mean and sample proportion areunbiased estimators of population mean andproportion. The following sample variance is anunbiased estimator of population variance.

1

)( 22

=

n

xxs

i


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Properties of

Point EstimatorsOf all the unbiased estimators, we preferthe estimator whose sampling distributionhas the smallest spread orvariability.


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An interval estimator draws inferencesabout a population by estimating the valueof an unknown parameter using an

interval.

Interval estimator

Population distribution

Sample distribution

Parameter

Interval Estimator


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Selecting the right sample statistic to estimatea parameter value depends on thecharacteristics of the statistic.

Estimators Characteristics

Estimators desirable characteristics:Unbiasedness:An unbiased estimator is one whose

expected value is equal to the parameter it estimates.Consistency:An unbiased estimator is said to be

consistent if the difference between the estimator and

the parameter grows smaller as the sample sizeincreases.

Relative eff iciency :For two unbiased estimators, the onewith a smaller variance is said to be relatively efficient.


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Uncertainty Analysis The estimate of the error is called the uncertainty.

It includes both bias and precision errors. We need to identify all the potential significant errors for

the instrument(s). All measurements should be given in three parts

Mean value

Uncertainty Confidence interval on which that uncertainty is based

(typically 95% C.I.) Uncertainty can be expressed in either

absolute terms (i.e., 5 Volts 0.5 Volts)or in percentage terms (i.e., 5 Volts 10%) We will use a confidence interval throughout this course.


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Estimating the Population Mean

when the Population Variance is

Known How is an interval estimator produced froma sampling distribution?

A sample of size n is drawn from thepopulation, and its mean is calculated.

By the central limit theorem, isnormally distributed (or approximatelynormally distributed.), thus

X

X


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n

x

Z s

m

=

We have established before that

=s

ms

m 1)n

zxn

z(P 22

Estimating the Population Mean when

the Population Variance is Known


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=sms 1)n

zxn

zx(P 22

This leads to the following equivalent statement

The Confidence Interval form( s is known)

The confidence interval


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CONFIDENCE INTERVAL FOR

KNOWN

Confidence Level :1 - = The probabilitythat a stated interval contains the unknownparameter.

m

s

/ 2X zn

s

/ 2X z

n

s

/ 2X z

n

s

Upper Confidence Limit

(UCL)

Lower Confidence Limit

(LCL)


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Interpreting the Confidence

Interval form1 of all the values of obtained in repeated

sampling from a given distribution, construct an interval

that includes (covers) the expected value of the

population.

x

s

s

n

zx,

n

zx 22


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x

nz2 2

s

n

zx 2s

n

zx 2s

Lower confidence limit Upper confidence limit

1 - Confidence level

Graphical Demonstration of the

Confidence Interval form


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Four commonly used confidence levels

Confidence

level /20.90 0.10 0.05 1.645

0.95 0.05 0.025 1.96

0.98 0.02 0.01 2.33

0.99 0.01 0.005 2.575

z/2


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Example: Estimate the mean value of the distribution resulting from

the throw of a fair die. It is known that s = 1.71. Use a 90%

confidence level, and 100 repeated throws of the die.

Solution: The confidence interval is


=s

n

zx 2 28.x100

71.1645.1x =

The mean values obtained in repeated draws of samples of size100 result in interval estimators of the form

[sample mean - .28, Sample mean + .28],

90% of which cover the real mean of the distribution.


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Recalculate the confidence interval for 95% confidence level.

Solution: =s

n

zx 2 34.x

100

71.196.1x =

34.x 34.x

.95

.90

28.x 28.x


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The width of the 90% confidence interval = 2(.28) = .56

The width of the 95% confidence interval = 2(.34) = .68

Because the 95% confidence interval is wider, it is

more likely to include the value ofm.


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Example Doll Computer Company delivers computers

directly to its customers who order via theInternet.

To reduce inventory costs in its warehouses Doll,employs an inventory model, that requires theestimate of the mean demand during lead time.

It is found that lead time demand is normallydistributed with a standard deviation of 75computers per lead time.

Estimate the lead time demand with 95%

confidence.

http://localhost/var/www/apps/conversion/tmp/scratch_12/Xm10-01.xlshttp://localhost/var/www/apps/conversion/tmp/scratch_12/Xm10-01.xls


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Example Solution

The parameter to be estimated is m, the meandemand during lead time.

We need to compute the interval estimation form.

From the data provided in the data file, thesample mean is


.16.370=x

56.399,76.34040.2916.37025

7596.116.370

25

75z16.370

nzx

025.2

===

=s

Since 1 - =.95, = .05.

Thus /2 = .025. Z.025 = 1.96


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Example (MINITAB Output)

Stat > Basic Statistics > 1-Sample Z

Z Confidence Intervals

The assumed sigma = 75.0

Variable N Mean StDev SE Mean 95.0 % CI

Demand 25 370.2 80.8 15.0 (340.8, 399.6)


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Example

To help make a decision about expansionplans, the president of a music companyneeds to know how many CDs teenagersbuy annually. 250 teenagers are selected;

each reports the number of CDs boughtwithin the last 12 months. The responseshave a sample mean of 4.26. The (population)standard deviation is 3. Construct 99, 95, 80

percent confidence intervals for the trueaverage number of CDs bought by teenagersin a year.


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ANSWER

99% CI:

95% CI: 80% CI:

. .z z

X Xn n

. .. .

. . %

005 005

2 57 3 2 57 34 26 4 26

250 250

3 771 4 749 99

s s m = m = m =

= 95%3.89 4.63m = 80%4.015 4.501m


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INTERPRETATION

3.8 as a guess regarding the average

number of CDs bought by teenagers is:

a) reasonable corresponding to 99%confidence

b) unreasonable corresponding to 95%confidence, and

c) unreasonable corresponding to 80%confidence.

I f ti d th Width f th


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Wide interval estimator provides littleinformation.

Where is m ????????????????

Information and the Width of the

Interval

I f ti d th Width f th


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Here is a much narrower interval.

If the confidence level remains

unchanged, the narrower interval

provides more meaningfulinformation.

Wide interval estimator provides littleinformation.Where is m ?

Ahaaa!

Information and the Width of the

Interval


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The width of the confidence interval isaffected by

the population standard deviation (s) the confidence level (1-) the sample size (n).

The Width of the Confidence Interval


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90%

Confidence level

To maintain a certain level of confidence, a larger

standard deviation requires a larger confidence interval.

n)645.1(2

nz2 05.

s=

s

/2 = .05/2 = .05

n

5.1

)645.1(2n

5.1

z2 05.s

=

s

Suppose the standard

deviation has increased

by 50%.

The Affects ofs on the interval width


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n)96.1(2

nz2

025.

s=

s

/2 = 2.5%/2 = 2.5%

/2 = 5%/2 = 5%

n

)645.1(2

n

z2 05.s

=s

Confidence level90%95%

Let us increase the

confidence level

from 90% to 95%.

Larger confidence level produces a wider confidence interval

The Affects of Changing the

Confidence Level

Th Aff t f Ch i th S l


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90%

Confidence level

n)645.1(2

nz2 05.

s=

s

Increasing the sample size decreases the width of the

confidence interval while the confidence level can remain

unchanged.

The Affects of Changing the Sample

Size


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The required sample size to estimate themean is

22

w

zn

s=

Selecting the Sample size


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Example To estimate the amount of lumber that

can be harvested in a tract of land, the

mean diameter of trees in the tractmust be estimated to within one inchwith 99% confidence.

What sample size should be taken?Assume that diameters are normallydistributed with s = 6 inches.



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Solution The estimate accuracy is +/-1 inch. That is w = 1.

The confidence level 99% leads to = .01, thusz

/2= z

.005= 2.575.

We compute239

1

)6(575.2

w

zn

22

2 =

=

s=

If the standard deviation is really 6 inches,the interval resulting from the random sampling

will be of the form . If the standard deviation

is greater than 6 inches the actual interval will

be wider than +/-1.

1x



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EXAMPLE

A sample survey of kindergarten children in NYcity is being planned to estimate, among theother things, the mean number of older siblingsof such children. It is desired to estimate this

mean within , with a 90% confidencelevel. A reasonable value for is 0.6.

a) What sample size is needed to estimate themean number of older siblings of such children?

w=0.08, s = 0.6,

0.08 s

0.05z 1.645=2

1.645* 0.6n 152.2 153

0.08

= =


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EXAMPLE(contd.)

b) If w=0.06, how will n change?

2

1.645* 0.6n 270.6 2710.06

= =

So, when w=0.08 0.06, the required sample

size increases substantially.


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Inference about the Population

Mean when the s is unknown When s is unknown, we need to estimate it to

construct confidence intervals for the population

mean s is estimated by s.

When s is unknown, the sample mean of a smallsample is no longer normally distributed. That is,we use t-scores (i.e., scores from a tdistribution) in place of z-scores.


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Inference About the PopulationMean when s is Unknown

The Student t Distribution

Standard Normal

Student t

0


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Effect of the Degrees of Freedom

on the t Density Function

Student t with 10 DF

0

Student t with 2 DF

Student t with 30 DF

The degrees of freedom, (a function of the sample size)

determine how spread the distribution is compared to the

normal distribution.

Finding t scores Under a


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Finding t-scores Under at-Distribution (t-tables)

t.100 t.05 t.025 t.01 t.005Degrees of

Freedom

1

2

3

45

6

7

8

910

11

12

3.078

1.886

1.638

1.5331.476

1.440

1.415

1.397

1.3831.372

1.363

1.356

6.314

2.920

2.353

2.1322.015

1.943

1.895

1.860

1.8331.812

1.796

1.782

12.706

4.303

3.182

2.7762.571

2.447

2.365

2.306

2.2622.228

2.201

2.179

31.821

6.965

4.541

3.7473.365

3.143

2.998

2.896

2.8212.764

2.718

2.681

63.657

9.925

5.841

4.6044.032

3.707

3.499

3.355

3.2503.169

3.106

3.055

t0.05, 10=1.812

1.812

.05

t0

C fid I t l


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Confidence Intervals on m(s unknown)

A (1100% confidence interval on m(when s is unknown) based on

samples of size n drawn from aNormal Population is given by:

/2, n-1 /2x t , (t with n-1 d.f.)

s

n

EXAMPLE


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EXAMPLE A new breakfast cereal is test-marked for

1 month at stores of a large supermarketchain. The result for a sample of16 storesindicate average sales of $1200 with a

sample standard deviation of $180. Set up99% confidence interval estimate of thetrue average sales of this new breakfast

cereal. Assume normality.

/ 2 ,n 1 0.005,15

n 16 ,x $1200,s $180, 0.01

t t 2.947

= = = =

= =


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ANSWER

99% CI form:

(1067.3985, 1332.6015)

With 99% confidence, the limits 1067.3985

and 1332.6015 cover the true averagesales of the new breakfast cereal.

/ 2 ,n 1

s 180x t 1200 2.947 1200 132.6015

n 16

= =


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Checking the required conditions

We need to check that the population isnormally distributed, or at least not extremelynonnormal.

There are statistical methods to test fornormality

From the sample histograms we see

0

5

10

15

20

25

30

-4 2 8 14 22 30 More


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Example

A random sample of n=8 E-glass fiber testspecimens of a certain type yielded asample mean interfacial shear yield

stress of 30.2 and a sample standarddeviation of 3.1. Assuming that interfacialshear yield stress is normally distributed,compute a 95% CI for true average stress.


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One Sided Confidence Intervals

An upper confidence bound form:

, 1n

s

x t nm

A lower confidence bound form:

, 1n

sx t

n

m


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Example

A sample of 14 joint specimens of aparticular type gave a sample meanproportional limit stress of 8.48 MPaand a sample standard deviation of 0.79MPa. Calculate and interpret a 95% lowerconfidence bound for the true averageproportional limit stress of all such joints.

What, if any, assumptions did you makeabout the distribution of proportional limitstress?

Estimation of a Population


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Estimation of a PopulationProportion

When the population consists of nominaldata, the only inference we can make isabout the proportion of occurrence of a

certain value. The parameter p was used before to

calculate these probabilities under the

binomial distribution.

Inference About a Population


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Inference About a PopulationProportion

Statistic and sampling distribution

the statistic used when making inference about p is:

Under certain conditions, [np > 5 and n(1-p) > 5],is approximately normally distributed, withm = p and s2= p(1 - p)/n.

.

.

xp where

nx the number of successes

n sample size

=


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Estimating the Proportion

Interval estimator for p (1- confidence level)

100(1-% Confidence Interval for p:

5)p1(nand5pnprov ided

n/)p1(pzp2/


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Nielsen Ratings

In a survey of2000 TV viewers at 11.40 p.m. on acertain night, 226indicated they watched TheTonight Show.

Estimate the number of TVs tuned to the Tonight

Show in a typical night, if there are 100 millionpotential television sets. Use a 95% confidencelevel.

Solution:

x=226, n=2000

014.113.

2000/)113.1(113.96.1113./)1( 2/

= nppzp

226 0.1132000

xpn = = =


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Solution

z - Estimate: ProportionViewers

Sample Proportion 0.113

Observations 2000

LCL 0.099

UCL 0.127

A confidence interval estimate of the

number of viewers who watched the

Tonight Show:

LCL = .099(100 million)= 9.9 millionUCL = .127(100 million)=12.7 million

Selecting the Sample Size to


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Selecting the Sample Size toEstimate the Proportion

Recall: The confidence interval for theproportion is

Thus, to estimate the proportion to within W,we can write

nppzp /)1( 2/

nppzW /)1(2/ =

Selecting the Sample Size to


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Selecting the Sample Size toEstimate the Proportion

The required sample size is

2

2/ )1(

=W

ppzn


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Example

Suppose we want to estimate the proportion ofcustomers who prefer our companys brand to

within .03 with 95% confidence. Find the sample size.

Solution:W = .03; 1 - = .95,

therefore /2 = .025,

so z.025 = 1.96

2

03.

)p1(p96.1n

=

Since the sample has not yetbeen taken, the sample proportion

is still unknown.

We proceed using either one of the

following two methods:

Sample Size to Estimate the

Proportion

Sample Size to Estimate the


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Method 1: There is no knowledge about the value of

Let . This results in the largest possible n needed fora 1- confidence interval of the form .

If the sample proportion does not equal .5, the actual W will

be narrower than .03 with the n obtained by the formulabelow.

5.p =03.p

p

068,103.

)5.1(5.96.1n

2

=

=

68303.

)2.1(2.96.1n

2

=

=

Sample Size to Estimate theProportion

Method 2: There is some idea about the value of

Use the value of to calculate the sample sizep

p

10. Parameter Estimation

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