Class 3 - CI

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Quantitative Method for Multi-dimensional Management and

Group Decision MakingEstimation of Population Parameters:

Confidence IntervalsClass 3

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Learning Objectives

1. State What Is Estimated

2. Distinguish Point & Interval Estimates

3. Explain Interval Estimates

4. Compute Confidence Interval Estimatesfor Population Mean & Proportion

5. Compute Sample Size

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Where Are We Going?

We are 95% confident that mean caloric intake of infants of low-income mothers receiving WICassistance is 80 to 200 kcal per day greater thanthat of infants of low-income mothers who do notreceive assistance. OR:

Infants of low-income mothers receiving WICassistance have a greater mean daily caloricintake than infants of low-income mothers notreceiving assistance (95%CI: 80 to 200 kcal).

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Thinking Challenge

Suppose youreinterested in theaverage amount of money thatstudents in this

class (thepopulation) have onthem. How wouldyou find out?

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Statistical Methods

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

Estimation HypothesisTesting

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Estimation Process

Mean, X , isunknown

Population Random SampleI am 95%

confident that X is between 40

& 60.

Mean X = 50

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Unknown PopulationParameters Are Estimated

Estimate PopulationParameter...

with SampleStatistic

Mean x x

Proportion p p s

Variance x 2 s 2

Differences 1 2 x 1 - x 2

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Estimation Methods

Estimation

PointEstimation

IntervalEstimation

ConfidenceInterval

PredictionInterval

Boot-strapping

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Estimation Methods

Estimation

PointEstimation

IntervalEstimation

ConfidenceInterval

PredictionInterval

Boot-strapping

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Interval Estimation

1. Provides Range of ValuesBased on Observations from 1 Sample

2. Gives Information about Closeness toUnknown Population Parameter

Stated in terms of Probability

Knowing Exact Closeness Requires KnowingUnknown Population Parameter

3. e.g., Unknown Population Mean LiesBetween 50 & 70 with 95% Confidence

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Key Elements ofInterval Estimation

Confidence Interval Sample Statistic(Point Estimate)

Confidence Limit(Lower)

Confidence Limit(Upper)

A Probability That the Population Parameter FallsSomewhere Within the Interval.

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Confidence Limitsfor Population Mean

( )

( )

1

5

x

x x

x

x x

x

x x

X Error

Error X X

Z X Error

Error Z

X Z

=

= +

=

=

=

=

(2) or

(3)

(4)

Parameter =Statistic Error

1984-1994 T/Maker Co.

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Many Samples HaveSame Confidence Interval

90% Samples

95% Samples

99% Samples

x+1.65 x x+2.58 x

x _

Xx+1.96 x

x-2.58 x x-1.65 xx-1.96 x

x

X = x Z x

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1. Probability that the UnknownPopulation Parameter Falls WithinInterval

2. Denoted (1 - ) % Is Probability That Parameter Is Not Within Interval

3. Typical Values Are 99%, 95%, 90%

Level of Confidence

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Intervals &Level of Confidence

SamplingDistribution

of Mean

Large Number of Intervals

Intervals

Extend from X - Z X to X + Z X

(1 - ) % of

IntervalsContain X . % Do Not.

x = x

1 - /2 /2

X _

x _

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Factors AffectingInterval Width

1. Data DispersionMeasured by X

2. Sample Size X = X / n

3. Level of Confidence(1 - )Affects Z

Intervals Extend from X - Z X to X + Z X

1984-1994 T/Maker Co.

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Confidence IntervalEstimates

ProportionMean

x Unknown

ConfidenceIntervals

Variance

FinitePopulationx Known

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Confidence Interval EstimateMean ( X Known)

1. AssumptionsPopulation Standard Deviation Is Known

Population Is Normally DistributedIf Not Normal, Can Be Approximated byNormal Distribution ( n 30)

2. Confidence Interval Estimate

X Z n

X Z n

X X

X +

/ /2 2

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Estimation ExampleMean ( X Known)

The mean of a random sample of n = 25is X = 50 . Set up a 95% confidenceinterval estimate for X if X = 10 .

X Z n

X Z n

X X

X

X

X

+

+

/ /

. .

. .

2 2

50 1961025

50 1961025

46 08 53 92

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Thinking Challenge

Youre a Q/C inspector for Gallo. The X for 2-liter

bottles is .05 liters. Arandom sample of 100 bottles showed X = 1.99liters. What is the 90% confidence intervalestimate of the true mean amount in 2-liter bottles?

2 liter

1984-1994 T/Maker Co.

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Confidence IntervalSolution*

X Z

n

X Z

n

X X

X

X

X

+

+

/ /

. ..

. ..

. .

2 2

199 164505

100

199 164505

1001982 1998

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ProportionMean

x Unknown

ConfidenceIntervals

Variance


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Confidence Interval EstimateMean ( X Unknown)

1. AssumptionsPopulation Standard Deviation Is Unknown

Population Must Be Normally Distributed

2. Use Students t Distribution

3. Confidence Interval Estimate X t

S

n X t

S

nn X n + / , / ,2 1 2 1

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Zt

Students t Distribution

0

t (df = 5)

StandardNormal

t (df = 13)Bell-ShapedSymmetric

Fatter Tails

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Upper Tail Area

df .25 .10 .05

1 1.000 3.078 6.314

2 0.817 1.886 2.9203 0.765 1.638 2.353

t0

Students t TableAssume:n = 3df = n - 1 = 2 = .10/2 =.05

2.920t Values

/ 2

.05

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Degrees of Freedom ( df )

1. Number of Observations that Are Freeto Vary After Sample Statistic Has

Been Calculated2. Example

Sum of 3 Numbers Is 6 X 1 = 1 (or Any Number) X 2 = 2 (or Any Number) X 3 = 3 (Cannot Vary)

Sum = 6

degrees of freedom= n -1= 3 -1= 2

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Estimation ExampleMean ( X Unknown)

A random sample of n = 25 has X = 50 &S = 8 . Set up a 95% confidence intervalestimate for X .

X t S

n X t

S

nn X n

X

X

+

+

/ , / ,

. .

. .

2 1 2 1

50 2 0639825

50 2 0639825

46 69 53 30

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Thinking Challenge

Youre a time studyanalyst in manufacturing.

Youve recorded thefollowing task times (min.):3.6, 4.2, 4.0, 3.5, 3.8, 3.1 .What is the 90% confidence intervalestimate of the populationmean task time?

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X = 3.7 S = 3.8987

n = 6, df = n -1 = 6 -1 = 5

S / n = 3.8987 / 6 = 1.592

t.05,5 = 2.01503.7 - (2.015)(1.592) X 3.7 + (2.015)(1.592)

.492 X 6.908

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ProportionMean

x Unknown

ConfidenceIntervals

Variance


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Confidence Interval EstimateProportion

1. AssumptionsTwo Categorical Outcomes

Population Follows Binomial DistributionNormal Approximation Can Be Used

n p 5 & n (1 - p ) 5

2. Confidence Interval Estimate

p Z p p

n p p Z

p p

nss s

ss s

+

( ) ( )1 1

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Estimation ExampleProportion

A random sample of 400 graduatesshowed 32 went to grad school. Set up a95% confidence interval estimate for p .

p Z p p

n p p Z

p p

n

p

p

ss s

ss s

+

+

/ /( ) ( )

. .. ( . )

. .. ( . )

. .

2 21 1

08 19608 1 08

40008 196

08 1 08

400

053 107

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Thinking Challenge

Youre a productionmanager for a newspaper.

You want to find the %defective. Of 200 newspapers, 35 haddefects. What is the

90% confidence intervalestimate of the populationproportion defective?

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n p 5n (1 - p ) 5

p Z p p

n p p Z p p

n

p

p

ss s

ss s

+

+

/ /( ) ( )

. .. (. )

. .. (. )

. .

2 2

1 1

175 1645175 825

200175 1645

175 825200

1308 2192

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Finding Sample Sizes

I dont want tosample too much

or too little!(1)

(2)

Z X Error

Error Z Z n

nZ

Error

x

x x

x x

x

=

=

= =

=

( )3

2 2

2

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Sample Size Example

What sample size is needed to be 90% confident of being correct within 5? Apilot study suggested that the standarddeviation is 45 .

n Z Error

x = = = 2 2

2

2 2

21645 45

5219 2 220 . .a fa f af

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Thinking Challenge

You work in HumanResources at Merrill Lynch.You plan to survey employeesto find their average medicalexpenses. You want to be95% confident that thesample mean is within $50 .

A pilot study showed that X was about $400 . What

sample size do you use?

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Sample SizeSolution*

nZ

Error

x =

=

=

2 2

2

2 2

2

196 400

50245 86 246

.

.

a fa f

a f

Class 3 - CI

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