Economics 173Business Statistics

Lecture 3

Fall, 2001

Professor J. Petry

http://www.cba.uiuc.edu/jpetry/Econ_173_fa01/

Introduction to Estimation

Introduction to Estimation

Chapter 9

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9.1 Introduction

• Statistical inference is the process by which we acquire information about populations from samples.

• There are two procedures for making inferences:– Estimation.– Hypotheses testing.

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9.2 Concepts of Estimation

• The objective of estimation is to determine the value of a population parameter on the basis of a sample statistic.

• There are two types of estimators– Point Estimator– Interval estimator

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– A point estimator draws inference about a population by estimating the value of an unknown parameter using a single value or a point.

Point estimator

Point Estimator

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Sample distribution

Point estimator

Population distributionParameter

?

• Point Estimator– A point estimator draws inference about a population

by estimating the value of an unknown parameter using a single value or a point.

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– An interval estimator draws inferences about a population by estimating the value of an unknown parameter using an interval.

– The interval estimator is affected by the sample size.

Interval estimator

Population distribution

Sample distribution

Parameter

Interval Estimator

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9.3 Estimating the Population Mean when the Population Standard Deviation is Known

• How is an interval estimator produced from a sampling distribution?– To estimate , a sample of size n is drawn from the

population, and its mean is calculated.– Under certain conditions, is normally distributed (or

approximately normally distributed.), thus

nx

Z

xx

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– We know that 1)( 22n

zxn

zP

1)n

zxn

zx(P 22

– This leads to the relationship

1 - of all the values of obtained in repeatedsampling from this distribution, construct an interval

that includes (covers) the expected value of thepopulation.

1 - of all the values of obtained in repeatedsampling from this distribution, construct an interval

that includes (covers) the expected value of thepopulation.

x

nzx,

nzx 22

10

x

nz2 2

nzx 2

n

zx 2

nzx,

nzx 22

Lower confidence limit Upper confidence limit

1 -

Confidence level

See simulation resultsdemonstrating this point

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0

50

100

150

Not all the confidence intervals cover the real expected value of 100.

1000

LCL

UCL

The selected confidence level is 90%,and 10 out of 100 intervals do not coverthe real

• The confidence interval are correct most, but not all, of the time.

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• Four commonly used confidence levelsConfidence

level 0.90 0.10 0.05 1.6450.95 0.05 0.025 1.960.98 0.02 0.01 2.330.99 0.01 0.005 2.575

Confidence level 0.90 0.10 0.05 1.6450.95 0.05 0.025 1.960.98 0.02 0.01 2.330.99 0.01 0.005 2.575

z

• Estimate the mean value of the distribution resulting from the throw of a fair die. It is known that = 1.71. Use 90% confidence level, and 100 repeated throws of the die.

• Solution: The confidence interval is

nzx 2

28.x10071.1

645.1x

The mean values obtained in repeated draws of samples of size 100 result in interval estimators of the form [sample mean - .28, Sample mean + .28]90% of which cover the real mean of the distribution.

The mean values obtained in repeated draws of samples of size 100 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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• 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 of

• 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 of

• Recalculate the confidence interval for 95% confidence level. • Solution:

n

zx 2 34.x10071.1

96.1x

34.x 34.x

.95

.90

28.x 28.x

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• Example 9.1– The number and the types of television programs and

commercials targeted at children is affected by the amount of time children watch TV.

– A survey was conducted among 100 North American children, in which they were asked to record the number of hours they watched TV per week.

– The population standard deviation of TV watch was known to be = 8.0

– Estimate the watch time with 95% confidence level.

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– The parameter to be estimated is the mean time of TV watch per week per child (of all American Children).

– We need to compute the interval estimator for – From the data in the book, the sample mean is:

761.28,621.2557.1191.27100

0.896.1191.27

100

0.8z191.27

nzx 025.2

.191.27x Since 1 - =.95, = .05. Thus /2 = .025. Z.025 = 1.96

• Solution

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• Interpreting the interval estimate– It is wrongwrong to state that the interval estimator is an

interval for which there is 1 - chance that the population mean lies between the LCL and the UCL.

– This is so because the is a parameter, not a random variable.

• Interpreting the interval estimate– It is wrongwrong to state that the interval estimator is an

interval for which there is 1 - chance that the population mean lies between the LCL and the UCL.

– This is so because the is a parameter, not a random variable.

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– LCL, UCL and the sample mean are the random variables, is a parameter, NOT a random variable.

– Thus, it is correct to state that there is 1 - chance that LCL will be less than and UCL will be greater than

nzxLCL 2

n

zxUCL 2

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• Example 9.2– To lower inventory costs, the Doll Computer

company wants to employ an inventory model.– Lead time demand is normally distributed with

standard deviation of 50 computers.– It is required to know the mean in order to calculate

optimum inventory levels.– Estimate the mean demand during lead time with

95% confidence.

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• Solution– The parameter to be estimated is

The interval estimator is

• Demand during 60 lead times is recorded514, 525, …., 476.

• The sample mean is calculated

• The 95% confidence interval is:

nzx,

nzx 22

75.49960985,29

n

xx i

4.512,1.48765.1275.49960

5096.175.499

nzx 2

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– Wide interval estimator provides little information.Where is

???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? ???? Ahaaa!

Here is a much smaller interval.If the confidence level remains unchanged, the smaller interval provides more meaningful information.

Here is a much smaller interval.If the confidence level remains unchanged, the smaller interval provides more meaningful information.

Information and the Width of the Interval

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The width of the interval estimate is a function of:• the population standard deviation• the confidence level• the sample size.

The width of the interval estimate is a function of:• the population standard deviation• the confidence level• the sample size.

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90%

Confidence level

To maintain a certain level of confidence,changing to a larger standard deviation requires a longer confidence interval.

To maintain a certain level of confidence,changing to a larger standard deviation requires a longer confidence interval.

n)645.1(2

nz2 05.

/2 = .05/2 = .05

n

5.1)645.1(2

n

5.1z2 05.

Suppose the standard deviationhas increased by 50%.

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90%

Confidence level95%

Let us increase the confidence level from 90% to 95%.

n)96.1(2

nz2 025.

/2 = 2.5%/2 = 2.5%

Increasing the sample size decreases the width of the interval estimate while the confidence level can remain unchanged.

Increasing the sample size decreases the width of the interval estimate while the confidence level can remain unchanged.

There is an inverserelationship between the width of the intervaland the sample size

Increasing the confidence level produces a wider interval

Increasing the confidence level produces a wider interval

n)645.1(2

nz2 05.

/2 = 5%/2 = 5%

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9.4 Selecting the Sample size

• We can control the width of the interval estimate by changing the sample size.

• Thus, we determine the interval width first, and derive the required sample size.

• The phrase “estimate the mean to within W units”, translates to an interval estimate of the form

wx wx

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• The required sample size to estimate the mean is

• Example 9.3– To estimate the amount of lumber that can be

harvested in a tract of land, the mean diameter of trees in the tract must be estimated to within one inch with 99% confidence.

– What sample size should be taken? (assume diameters are normally distributed with = 6 inches.

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w

zn

22

w

zn

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• Solution– The estimate accuracy is +/-1 inch. That is w = 1.– The confidence level 99% leads to = .01, thus z/2

= z.005 = 2.575.– We compute 239

1)6(575.2

w

zn

222