8 - 1 © 1998 Prentice-Hall, Inc. Chapter 8 Inferences Based on a Single Sample: Tests of...

8 - 8 - 11

© 1998 Prentice-Hall, Inc.© 1998 Prentice-Hall, Inc.

Chapter 8Chapter 8Inferences Based on a Single Inferences Based on a Single Sample: Tests of HypothesisSample: Tests of Hypothesis

8 - 8 - 22


Learning ObjectivesLearning Objectives

1.1. Distinguish types of hypotheses Distinguish types of hypotheses

2.2. Describe hypothesis testing processDescribe hypothesis testing process

3.3. Explain Explain pp-value concept-value concept

4.4. Solve hypothesis testing problems Solve hypothesis testing problems based on a single samplebased on a single sample

8 - 8 - 33


Types of Types of Statistical Statistical

ApplicationsApplications

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

8 - 8 - 44


Hypothesis Testing Hypothesis Testing ConceptsConcepts

8 - 8 - 55


Hypothesis TestingHypothesis Testing

8 - 8 - 66



PopulationPopulation

8 - 8 - 77




I believe the population mean age is 50 (hypothesis).

8 - 8 - 88





MeanMean X X = 20= 20

Random Random samplesample

8 - 8 - 99





MeanMean X X = 20= 20

Reject hypothesis! Not close.

Reject hypothesis! Not close.

Random Random samplesample

8 - 8 - 1010


What’s a What’s a Hypothesis?Hypothesis?

1.1. A belief about a A belief about a population parameterpopulation parameter Parameter is Parameter is

populationpopulation mean, mean, proportion, varianceproportion, variance

Must be statedMust be statedbeforebefore analysis analysis

8 - 8 - 1111


What’s a What’s a Hypothesis?Hypothesis?

1.1. A belief about a A belief about a population parameterpopulation parameter Parameter is Parameter is

populationpopulation mean, mean, proportion, varianceproportion, variance

Must be statedMust be statedbeforebefore analysis analysis

I believe the mean GPA I believe the mean GPA of this class is 3.5!of this class is 3.5!

© 1984-1994 T/Maker Co.

8 - 8 - 1212


Null HypothesisNull Hypothesis

1.1. What is testedWhat is tested

2.2. Has serious outcome if incorrect decision Has serious outcome if incorrect decision mademade

3.3. Always has equality sign: Always has equality sign: , or , or 4.4. Designated HDesignated H00

5.5. Specified as HSpecified as H00: : Some numeric value Some numeric value Written with = sign even if Written with = sign even if , or , or Example, HExample, H00: : 3 3

8 - 8 - 1313


Alternative Alternative HypothesisHypothesis

1.1. Opposite of null hypothesisOpposite of null hypothesis

2.2. Always has inequality sign:Always has inequality sign: ,,, or , or

3.3. Designated HDesignated Haa

4.4. Specified HSpecified Haa: : < Some value < Some value Example, HExample, Haa: : < 3 < 3

8 - 8 - 1414


Identifying Identifying HypothesesHypotheses

StepsSteps

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StepsStepsStepsSteps

1.1. State question statisticallyState question statistically

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ExampleExample

Is the population mean Is the population mean different from 3?different from 3?

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ExampleExample


1.1. 3 3

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2.2. State opposite statisticallyState opposite statistically Must be mutually exclusive Must be mutually exclusive

& exhaustive& exhaustive

ExampleExample


1.1. 3 3

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ExampleExample


1.1. 3 3

2.2. = 3 = 3

8 - 8 - 2020







3.3. Select & state alternative Select & state alternative hypothesishypothesis Has the Has the , , <<, or , or > > signsign

ExampleExample


1.1. 3 3

2.2. = 3 = 3

8 - 8 - 2121








ExampleExample


1.1. 3 3

2.2. = 3 = 3

3.3. H Haa: : 3 3

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4.4. State null hypothesisState null hypothesis

ExampleExample


1.1. 3 3

2.2. = 3 = 3

3.3. H Haa: : 3 3

8 - 8 - 2323









ExampleExample


1.1. 3 3

2.2. = 3 = 3

3.3. H Haa: : 3 3

4.4. H H00: : = 3 = 3

8 - 8 - 2424


Identifying Identifying HypothesesHypothesesExample 1Example 1

StepsSteps






ExampleExample

Is the population Is the population average amount of TV average amount of TV viewing 12 hours? viewing 12 hours?

1.1.

2.2.

3.3.

4.4.

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StepsSteps






ExampleExample


1.1. = 12 = 12

2.2.

3.3.

4.4.

8 - 8 - 2626



StepsSteps






ExampleExample


1.1. = 12 = 12

2.2. 12 12

3.3.

4.4.

8 - 8 - 2727



StepsSteps






ExampleExample


1.1. = 12 = 12

2.2. 12 12

3.3. H Haa: : 12 12

4.4.

8 - 8 - 2828



StepsSteps






ExampleExample


1.1. = 12 = 12

2.2. 12 12

3.3. H Haa: : 12 12

4.4. H H00: : = 12 = 12

8 - 8 - 2929



StepsSteps






ExampleExample

Is the population Is the population average amount of TV average amount of TV viewing viewing differentdifferent from from 12 hours? 12 hours?

1.1.

2.2.

3.3.

4.4.

8 - 8 - 3030



StepsSteps






ExampleExample


1.1. 12 12

2.2.

3.3.

4.4.

8 - 8 - 3131



StepsSteps






ExampleExample


1.1. 12 12

2.2. = 12= 12

3.3.

4.4.

8 - 8 - 3232



StepsSteps






ExampleExample


1.1. 12 12

2.2. = 12= 12

3.3. H Haa: : 12 12

4.4.

8 - 8 - 3333



StepsSteps






ExampleExample


1.1. 12 12

2.2. = 12= 12

3.3. H Haa: : 12 12

4.4. H H00: : = 12 = 12

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StepsSteps






ExampleExample

Is the average cost Is the average cost per hat less than or per hat less than or equal to $20?equal to $20?

1.1.

2.2.

3.3.

4.4.

8 - 8 - 3535



StepsSteps






ExampleExample


1.1. 20 20

2.2.

3.3.

4.4.

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StepsSteps






ExampleExample


1.1. 20 20

2.2. 20 20

3.3.

4.4.

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StepsSteps






ExampleExample


1.1. 20 20

2.2. 20 20

3.3. H Haa: : 20 20

4.4.

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StepsSteps






ExampleExample


1.1. 20 20

2.2. 20 20

3.3. H Haa: : 20 20

4.4. H H00: : = 20 = 20

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StepsSteps






ExampleExample

Is the average amount Is the average amount spent in the bookstore spent in the bookstore greater than $25?greater than $25?

1.1.

2.2.

3.3.

4.4.

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StepsSteps






ExampleExample


1.1. 25 25

2.2.

3.3.

4.4.

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StepsSteps






ExampleExample


1.1. 25 25

2.2. 25 25

3.3.

4.4.

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StepsSteps






ExampleExample


1.1. 25 25

2.2. 25 25

3.3. H Haa: : 25 25

4.4.

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StepsSteps






ExampleExample


1.1. 25 25

2.2. 25 25

3.3. H Haa: : 25 25

4.4. H H00: : = 25 = 25

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Basic IdeaBasic Idea

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Sample Mean = 50 Sample Mean = 50

HH00HH00

Sampling DistributionSampling Distribution

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It is unlikely It is unlikely that we would that we would get a sample get a sample mean of this mean of this value ...value ...

20202020HH00HH00

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... if in fact this were... if in fact this were the population mean the population mean

20202020HH00HH00

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... if in fact this were... if in fact this were the population mean the population mean

... therefore, ... therefore, we reject the we reject the hypothesis hypothesis

that that = 50.= 50.

20202020HH00HH00

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Level of SignificanceLevel of Significance

1.1. Defines unlikely values of sample statistic Defines unlikely values of sample statistic if null hypothesis is trueif null hypothesis is true Called rejection region of sampling Called rejection region of sampling

distributiondistribution

2.2. Is a probability Is a probability

3.3. Denoted Denoted (alpha)(alpha)

4.4. Selected by researcher at startSelected by researcher at start Typical values are .01, .05, .10Typical values are .01, .05, .10

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Rejection Region Rejection Region (One-Tail Test) (One-Tail Test)

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HoHoValueValue

Sample StatisticSample Statistic



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HoHoValueValue


RejectionRejectionRegionRegion



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HoHoValueValue



NonrejectionNonrejectionRegionRegion



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HoHoValueValue






CriticalCriticalValueValue

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HoHoValueValue






CriticalCriticalValueValue

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HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -

Level of ConfidenceLevel of Confidence

8 - 8 - 5757



HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -


Observed sample statisticObserved sample statistic

8 - 8 - 5858



HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -


8 - 8 - 5959


Rejection Regions Rejection Regions (Two-Tailed Test) (Two-Tailed Test)

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HoHoValueValue




8 - 8 - 6161


HoHoValueValue






8 - 8 - 6262


HoHoValueValue







8 - 8 - 6363


HoHoValueValue CriticalCritical

ValueValueCriticalCriticalValueValue







8 - 8 - 6464


HoHoValueValue CriticalCritical

ValueValueCriticalCriticalValueValue

1/2 1/2 1/2 1/2







8 - 8 - 6565



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


8 - 8 - 6666



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


8 - 8 - 6767



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


8 - 8 - 6868



HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -


8 - 8 - 6969


Decision Making RisksDecision Making Risks

8 - 8 - 7070


Errors in Errors in Making DecisionMaking Decision

1.1. Type I errorType I error Reject true null hypothesisReject true null hypothesis Has serious consequencesHas serious consequences Probability of Type I error is Probability of Type I error is (alpha)(alpha)

Called level of significanceCalled level of significance

2.2. Type II errorType II error Do not reject false null hypothesisDo not reject false null hypothesis Probability of Type II error is Probability of Type II error is (beta)(beta)

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Jury Trial H0 Test

Actual Situation Actual Situation

Verdict Innocent Guilty Decision H0 True H0

False

Innocent Correct ErrorDo NotReject

H0

1 - Type IIError

()

Guilty Error Correct RejectH0

Type IError ()

Power(1 - )

Jury Trial H0 Test



False


H0

1 - Type IIError

()


Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

8 - 8 - 7272


Jury Trial H0 Test



False


H0

1 - Type IIError

()


Type IError ()

Power(1 - )

Jury Trial H0 Test



False


H0

1 - Type IIError

()


Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

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Hypothesis Testing Hypothesis Testing StepsSteps

8 - 8 - 7474


HH00 Testing Steps Testing Steps

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State HState H00

State HState H11

Choose Choose

Choose Choose nn

Choose testChoose test

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Set up critical valuesSet up critical values

Collect dataCollect data

Compute test statisticCompute test statistic

Make statistical decisionMake statistical decision

Express decisionExpress decision

State HState H00

State HState H11

Choose Choose

Choose Choose nn

Choose testChoose test

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One Population One Population TestsTests

8 - 8 - 7878



OnePopulation

8 - 8 - 7979



OnePopulation

Mean

8 - 8 - 8080



OnePopulation

Mean Proportion

8 - 8 - 8181



OnePopulation

Z Test(1 & 2tail)

Mean ProportionLargeLargeLargeLargeSampleSampleSampleSample

8 - 8 - 8282



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Mean ProportionLargeLargeLargeLargeSampleSampleSampleSample

SmallSmallSmallSmallSampleSampleSampleSample

8 - 8 - 8383



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

LargeSample

Z Test(1 & 2tail)

Mean ProportionSmallSample

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

LargeSample

Z Test(1 & 2tail)


8 - 8 - 8484


Two-Tailed Z Test Two-Tailed Z Test of Mean (Large Sample)of Mean (Large Sample)

8 - 8 - 8585



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

LargeSample

Z Test(1 & 2tail)


OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

LargeSample

Z Test(1 & 2tail)


8 - 8 - 8686


Two-Tailed Z Test Two-Tailed Z Test for Mean (Large for Mean (Large

Sample)Sample)1.1. AssumptionsAssumptions

Sample size at least 30 (Sample size at least 30 (nn 30) 30) If population standard deviation unknown, If population standard deviation unknown,

use sample standard deviationuse sample standard deviation

2.2. Alternative hypothesis has Alternative hypothesis has sign sign

8 - 8 - 8787


Two-Tailed Z Test Two-Tailed Z Test for Mean (Large for Mean (Large




2.2. Alternative hypothesis has Alternative hypothesis has sign sign

3.3. Z-test statisticZ-test statistic

ZX X

n

x

x

Z

X X

n

x

x

8 - 8 - 8888


Two-Tailed Z TestTwo-Tailed Z Test Example Example

Does an average box of Does an average box of cereal contain cereal contain 368368 grams grams of cereal? A random of cereal? A random sample of sample of 3636 boxes boxes showedshowedX = 372.5X = 372.5. The . The company has specified company has specified to be to be 2525 grams. Test at grams. Test at the the .05.05 level. level. 368 gm.368 gm.

8 - 8 - 8989


Two-Tailed Z Test Two-Tailed Z Test SolutionSolution

HH00: :

HHaa: :

nn

Critical Value(s):Critical Value(s):

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

8 - 8 - 9090



HH00: : = 368 = 368

HHaa: : 368 368

nn



Decision:Decision:


8 - 8 - 9191



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 3636



Decision:Decision:


8 - 8 - 9292



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 3636



Decision:Decision:


Z0 1.96-1.96

.025

Reject H 0 Reject H 0

.025

Z0 1.96-1.96

.025


.025

8 - 8 - 9393



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

8 - 8 - 9494



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

Do not reject at Do not reject at = .05 = .05

ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

8 - 8 - 9595



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025


No evidence No evidence average is not 368average is not 368

ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

8 - 8 - 9696


Two-Tailed Z Test Two-Tailed Z Test Thinking ChallengeThinking Challenge

You’re a Q/C inspector. You want to You’re a Q/C inspector. You want to find out if a new machine is making find out if a new machine is making electrical cords to customer electrical cords to customer specification: specification: averageaverage breaking breaking strength of strength of 7070 lb. with lb. with = 3.5 = 3.5 lb. lb. You take a sample of You take a sample of 3636 cords & cords & compute a sample mean of compute a sample mean of 69.769.7 lb. lb. At the At the .05.05 level, is there evidence level, is there evidence that the machine is that the machine is notnot meeting the meeting the average breaking strength?average breaking strength?

AloneAlone GroupGroup Class Class

8 - 8 - 9797


Two-Tailed Z Test Two-Tailed Z Test Solution*Solution*

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:


8 - 8 - 9898



HH00: : = 70 = 70

HHaa: : 70 70

= =

nn = =



Decision:Decision:


8 - 8 - 9999



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


8 - 8 - 100100



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

8 - 8 - 101101



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

69 7 70

3 536

51.

..Z

X

n

69 7 70

3 536

51.

..

8 - 8 - 102102



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

69 7 70

3 536

51.

..Z

X

n

69 7 70

3 536

51.

..


8 - 8 - 103103



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

69 7 70

3 536

51.

..Z

X

n

69 7 70

3 536

51.

..


No evidence No evidence average is not 70average is not 70

8 - 8 - 104104


One-Tailed Z Test One-Tailed Z Test of Mean (Large Sample)of Mean (Large Sample)

8 - 8 - 105105


One-Tailed Z Test One-Tailed Z Test for Mean (Large for Mean (Large




2.2. Alternative hypothesis has < or > signAlternative hypothesis has < or > sign

8 - 8 - 106106


One-Tailed Z Test One-Tailed Z Test for Mean (Large for Mean (Large




2.2. Alternative hypothesis has Alternative hypothesis has or > signor > sign

3.3. Z-test statisticZ-test statistic

ZX X

n

x

x

Z

X X

n

x

x

8 - 8 - 107107


One-Tailed Z Test One-Tailed Z Test for Mean Hypothesesfor Mean Hypotheses

8 - 8 - 108108


Z0

Reject H 0

Z0

Reject H 0


HH00::==0 H0 Haa: : << 0 0

Must be Must be significantlysignificantly below below

8 - 8 - 109109


Z0

Reject H 0

Z0

Reject H 0

Z0

Reject H 0

Z0

Reject H 0


HH00::==0 H0 Haa: : << 0 0 HH00::==0 H0 Haa: : >> 0 0

Must be Must be significantlysignificantly below below

Small values satisfy Small values satisfy HH0 0 . Don’t reject!. Don’t reject!

8 - 8 - 110110


One-Tailed Z Test One-Tailed Z Test Finding Critical ZFinding Critical Z

8 - 8 - 111111


Z0

= 1

Z0

= 1


What is Z given What is Z given = .025? = .025?

= .025= .025

8 - 8 - 112112


Z0

= 1

Z0

= 1


.500 .500 -- .025.025

.475.475


= .025= .025

8 - 8 - 113113


Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

Z0

= 1

Z0

= 1


.500 .500 -- .025.025

.475.475

.06

1.9 .4750.4750

Standardized Normal Standardized Normal Probability Table (Portion)Probability Table (Portion)


= .025= .025

8 - 8 - 114114


Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

Z0

= 1

1.96 Z0

= 1

1.96


.500 .500 -- .025.025

.475.475.06.06

1.91.9 .4750

Standardized Normal Standardized Normal Probability Table (Portion)Probability Table (Portion)


= .025= .025

8 - 8 - 115115


One-Tailed Z TestOne-Tailed Z Test Example Example

Does an average box of Does an average box of cereal contain cereal contain more thanmore than 368368 grams of cereal? A grams of cereal? A random sample of random sample of 36 36 boxes showedboxes showedX = 372.5X = 372.5. . The company has The company has specified specified to be to be 2525 grams. Test at the grams. Test at the .05.05 level.level.

368 gm.368 gm.

8 - 8 - 116116


One-Tailed Z Test One-Tailed Z Test SolutionSolution

HH00: :

HHaa: :

= =

n n = =



Decision:Decision:


8 - 8 - 117117



HH00: : = 368 = 368

HHaa: : > 368 > 368

= =

n n = =



Decision:Decision:


8 - 8 - 118118



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 3636



Decision:Decision:


8 - 8 - 119119



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 3636



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject

8 - 8 - 120120



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject

ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

8 - 8 - 121121



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject Reject at Reject at = .05 = .05

ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

8 - 8 - 122122



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05


Evidence average is Evidence average is more than 368more than 368

ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

8 - 8 - 123123


One-Tailed Z Test One-Tailed Z Test Thinking ChallengeThinking Challenge

You’re an analyst for Ford. You You’re an analyst for Ford. You want to find out if the average want to find out if the average miles per gallon of Escorts is at miles per gallon of Escorts is at least 32 mpg. Similar models least 32 mpg. Similar models have a standard deviation of have a standard deviation of 3.83.8 mpg. You take a sample of mpg. You take a sample of 6060 Escorts & compute a sample Escorts & compute a sample mean of mean of 30.730.7 mpg. At the mpg. At the .01.01 level, is there evidence that the level, is there evidence that the miles per gallon is miles per gallon is at leastat least 3232??


8 - 8 - 124124


One-Tailed Z Test One-Tailed Z Test Solution*Solution*

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:



Decision:Decision:


8 - 8 - 125125



HH00: : = 32 = 32

HHaa: : < 32 < 32

= =

nn = =



Decision:Decision:



Decision:Decision:


8 - 8 - 126126



HH00: : = 32 = 32

HHaa: : < 32 < 32

== .01 .01

nn = = 6060



Decision:Decision:



Decision:Decision:


8 - 8 - 127127



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

8 - 8 - 128128



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..Z

X

n

30 7 32

3 860

2 65.

..

8 - 8 - 129129



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..Z

X

n

30 7 32

3 860

2 65.

..

Reject at Reject at = .01 = .01

8 - 8 - 130130



HH00: : = 32 = 32

HHaa: : < 32 < 32

= .01= .01

nn = 60= 60



Decision:Decision:



Decision:Decision:


Z0-2.33

.01

Reject

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..Z

X

n

30 7 32

3 860

2 65.

..


There is evidence There is evidence average is less than 32average is less than 32

8 - 8 - 131131


Observed Significance Observed Significance Levels: Levels: pp-Values-Values

8 - 8 - 132132


pp-Value-Value

1.1. Probability of obtaining a test statistic more Probability of obtaining a test statistic more extreme (extreme (or or than the actual sample than the actual sample value given Hvalue given H00 is true is true

2.2. Called observed level of significanceCalled observed level of significance Smallest value of Smallest value of H H00 can be rejected can be rejected

3.3. Used to make rejection decisionUsed to make rejection decision If If pp-value -value , do not reject H, do not reject H00

If If pp-value < -value < , reject H, reject H00

8 - 8 - 133133


Two-Tailed Z Test Two-Tailed Z Test pp-Value Example -Value Example

Does an average box of Does an average box of cereal contain cereal contain 368368 grams grams of cereal? A random of cereal? A random sample of sample of 3636 boxes boxes showedshowedX = 372.5X = 372.5. The . The company has specified company has specified to be to be 2525 grams. Find the grams. Find the pp-value.-value. 368 gm.368 gm.

8 - 8 - 134134


Two-Tailed Z Test Two-Tailed Z Test pp-Value Solution-Value Solution

8 - 8 - 135135



Z0 1.80-1.80 Z0 1.80-1.80

Z value of sample Z value of sample statistic (observed)statistic (observed)

ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

8 - 8 - 136136




pp-value = -value = PP(Z (Z -1.80 or Z -1.80 or Z 1.80) 1.80)

Z0 1.80-1.80 Z0 1.80-1.80

8 - 8 - 137137



Z0 1.80-1.80

1/2 p-value1/2 p-value

Z0 1.80-1.80




8 - 8 - 138138


Z0 1.80-1.80


Z0 1.80-1.80




From Z table: From Z table: lookup 1.80lookup 1.80

.4641.4641


8 - 8 - 139139


Z0 1.80-1.80


Z0 1.80-1.80





.4641.4641

.5000.5000-- .4641.4641

.0359.0359


8 - 8 - 140140


Z0 1.80-1.80

1/2 p-value.0359

1/2 p-value.0359

Z0 1.80-1.80

1/2 p-value.0359

1/2 p-value.0359


pp-value = -value = PP(Z (Z -1.80 or Z -1.80 or Z 1.80) = 1.80) = .0718.0718

Z value of sample Z value of sample statisticstatistic


.4641.4641

.5000.5000-- .4641.4641

.0359.0359

8 - 8 - 141141



0 1.80-1.80 Z

RejectReject

0 1.80-1.80 Z

RejectReject

1/2 p-value = .03591/2 p-value = .03591/2 p-value = .03591/2 p-value = .0359

1/2 1/2 = .025 = .0251/2 1/2 = .025 = .025

8 - 8 - 142142



0 1.80-1.80 Z

RejectReject

0 1.80-1.80 Z

RejectReject

1/2 p-value = .03591/2 p-value = .03591/2 p-value = .03591/2 p-value = .0359

1/2 1/2 = .025 = .0251/2 1/2 = .025 = .025

((pp-value = .0718) -value = .0718) ( ( = .05). Do not = .05). Do not reject.reject.

8 - 8 - 143143


One-Tailed Z Test One-Tailed Z Test pp-Value Example -Value Example

Does an average box of Does an average box of cereal contain cereal contain more thanmore than 368368 grams of cereal? A grams of cereal? A random sample of random sample of 3636 boxes showedboxes showedX = 372.5X = 372.5. . The company has The company has specified specified to be to be 2525 grams. Find the grams. Find the pp-value.-value. 368 gm.368 gm.

8 - 8 - 144144


One-Tailed Z Test One-Tailed Z Test pp-Value Solution-Value Solution

8 - 8 - 145145



ZZ00

8 - 8 - 146146



Use Use alternative alternative hypothesis hypothesis to find to find directiondirection ZZ00

p-valuep-value

8 - 8 - 147147



Z0 1.80

p-value

Z0 1.80

p-valueUse Use alternative alternative hypothesis hypothesis to find to find directiondirection


ZX

n

372 5 368

1536

180.

.ZX

n

372 5 368

1536

180.

.

1.801.80

8 - 8 - 148148



Z0 1.80

p-value

Z0 1.80



pp-value is -value is PP(Z (Z 1.80) 1.80)

8 - 8 - 149149


Z0 1.80

p-value

Z0 1.80

p-value


Use Use alternative alternative hypothesis hypothesis to find to find directiondirection




.4641.4641

8 - 8 - 150150


Z0 1.80

p-value

Z0 1.80

p-value

.4641.4641






.5000.5000-- .4641.4641

.0359.0359

8 - 8 - 151151


Z0 1.80

p-value

Z0 1.80

p-value

.4641.4641





.5000.5000-- .4641.4641

.0359.0359

pp-value is -value is PP(Z (Z 1.80) = 1.80) = .0359.0359

8 - 8 - 152152


ZZ00 1.801.80

p-valuep-value


= .0359= .0359

8 - 8 - 153153


ZZ00 1.801.80

p-valuep-value


= .0359= .0359RejectReject

= .05= .05

8 - 8 - 154154


ZZ00 1.801.80

p-valuep-value



= .05= .05

8 - 8 - 155155


((pp-value = .0359) -value = .0359) ( ( = .05). Reject. = .05). Reject.

Z0 1.80

p-value

Z0 1.80

p-value



= .05= .05

8 - 8 - 156156


pp-Value -Value Thinking ChallengeThinking Challenge

You’re an analyst for Ford. You You’re an analyst for Ford. You want to find out if the average want to find out if the average miles per gallon of Escorts is miles per gallon of Escorts is at at least 32 least 32 mpg. Similar models mpg. Similar models have a standard deviation of have a standard deviation of 3.83.8 mpg. You take a sample of mpg. You take a sample of 6060 Escorts & compute a sample Escorts & compute a sample mean of mean of 30.730.7 mpg. What is the mpg. What is the value of the observed level of value of the observed level of significance (significance (pp-value-value)?)?


8 - 8 - 157157


pp-Value -Value Solution*Solution*

ZZ00

8 - 8 - 158158



Z0

p-value

Z0


8 - 8 - 159159



Z0-2.65

p-value

Z0-2.65

p-value

Z value of Z value of sample statisticsample statistic


ZX

n

30 7 32

3 860

2 65.

..Z

X

n

30 7 32

3 860

2 65.

..

8 - 8 - 160160



Z0-2.65

p-value

Z0-2.65

p-value



.4960.4960


8 - 8 - 161161



Z0-2.65

p-Value.004

Z0-2.65

p-Value.004



.4960.4960


.5000.5000-- .4960.4960

.0040.0040

8 - 8 - 162162



Z0-2.65

p-Value.004

Z0-2.65

p-Value.004



.4960.4960


.5000.5000-- .4960.4960

.0040.0040

pp-value = -value = PP(Z (Z -2.65) = .004. -2.65) = .004.pp-value < (-value < ( = .01). Reject H = .01). Reject H00..

8 - 8 - 163163


Two-Tailed t Test Two-Tailed t Test of Mean (Small Sample)of Mean (Small Sample)

8 - 8 - 164164



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

LargeSample

Z Test(1 & 2tail)


OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

LargeSample

Z Test(1 & 2tail)


8 - 8 - 165165


t Test for Mean t Test for Mean (Small Sample)(Small Sample)

1.1. AssumptionsAssumptions Sample size is less than 30 (n < 30)Sample size is less than 30 (n < 30) Population is normally distributedPopulation is normally distributed Population standard deviation is unknownPopulation standard deviation is unknown

8 - 8 - 166166


t Test for Mean t Test for Mean (Small Sample)(Small Sample)

1.1. AssumptionsAssumptions Sample size is less than 30 (n < 30)Sample size is less than 30 (n < 30) Population is normally distributedPopulation is normally distributed Population standard deviation is unknownPopulation standard deviation is unknown

3.3. T test statisticT test statistic

tX

Sn

tX

Sn

8 - 8 - 167167


Two-Tailed t TestTwo-Tailed t Test Finding Critical t Finding Critical t

ValuesValues

8 - 8 - 168168


t0 t0


ValuesValuesGiven: n = 3; Given: n = 3; = .10 = .10

8 - 8 - 169169


t0 t0


ValuesValues

/2 = .05/2 = .05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

8 - 8 - 170170


t0 t0


ValuesValues

/2 = .05/2 = .05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = 2df = n - 1 = 2

8 - 8 - 171171


v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182t0 t0


ValuesValuesCritical Values of t Table Critical Values of t Table

(Portion)(Portion)

/2 = /2 = .05.05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = df = n - 1 = 22

8 - 8 - 172172


v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

v t.10 t.05 t.025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182t0 2.920-2.920 t0 2.920-2.920


ValuesValuesCritical Values of t Table Critical Values of t Table

(Portion)(Portion)

/2 = .05/2 = .05

/2 = .05/2 = .05

Given: n = 3; Given: n = 3; = .10 = .10

df = n - 1 = 2df = n - 1 = 2

8 - 8 - 173173


Two-Tailed t TestTwo-Tailed t Test Example Example

Does an average box of Does an average box of cereal contain cereal contain 368368 grams of cereal? A grams of cereal? A random sample of random sample of 2525 boxes had a mean of boxes had a mean of 372.5372.5 & a standard & a standard deviation ofdeviation of 1212 grams. grams. Test at the Test at the .05.05 level. level. 368 gm.368 gm.

8 - 8 - 174174


Two-Tailed t Test Two-Tailed t Test SolutionSolution

HH00: :

HHaa: :

= =

df = df = Critical Value(s):Critical Value(s):


Decision:Decision:


8 - 8 - 175175



HH00: : = 368 = 368

HHaa: : 368 368

= =

df = df = Critical Value(s):Critical Value(s):


Decision:Decision:


8 - 8 - 176176



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24Critical Value(s):Critical Value(s):


Decision:Decision:


8 - 8 - 177177



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.064-2.064

.025


.025

t0 2.064-2.064

.025


.025

8 - 8 - 178178



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


tX

Sn

372 5 368

1225

1875.

.tX

Sn

372 5 368

1225

1875.

.

t0 2.064-2.064

.025


.025

t0 2.064-2.064

.025


.025

8 - 8 - 179179



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:



tX

Sn

372 5 368

1225

1875.

.tX

Sn

372 5 368

1225

1875.

.

t0 2.064-2.064

.025


.025

t0 2.064-2.064

.025


.025

8 - 8 - 180180



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:



There is no evidence There is no evidence pop. average is not 368pop. average is not 368

tX

Sn

372 5 368

1225

1875.

.tX

Sn

372 5 368

1225

1875.

.

t0 2.064-2.064

.025


.025

t0 2.064-2.064

.025


.025

8 - 8 - 181181


Two-Tailed t TestTwo-Tailed t TestThinking ChallengeThinking Challenge

You work for the FTC. A You work for the FTC. A manufacturer of detergent manufacturer of detergent claims that the mean weight claims that the mean weight of detergent is of detergent is 3.253.25 lb. You lb. You take a random sample of take a random sample of 1616 containers. You calculate the containers. You calculate the sample average to be sample average to be 3.2383.238 lb. with a standard deviation lb. with a standard deviation of of .117.117 lb. At the lb. At the .01.01 level, is level, is the manufacturer correct?the manufacturer correct?

3.25 lb.3.25 lb.


8 - 8 - 182182


Two-Tailed t Test Two-Tailed t Test Solution*Solution*

HH00: :

HHaa: :

df df



Decision:Decision:


8 - 8 - 183183



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

df df



Decision:Decision:


8 - 8 - 184184



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 16 - 1 = 1516 - 1 = 15



Decision:Decision:


8 - 8 - 185185



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 16 - 1 = 1516 - 1 = 15



Decision:Decision:


t0 2.947-2.947

.005


.005

t0 2.947-2.947

.005


.005

8 - 8 - 186186



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 16 - 1 = 1516 - 1 = 15



Decision:Decision:


tX

Sn

3 238 3 25

11764

82. .

..t

XSn

3 238 3 25

11764

82. .

..

t0 2.947-2.947

.005


.005

t0 2.947-2.947

.005


.005

8 - 8 - 187187



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 16 - 1 = 1516 - 1 = 15



Decision:Decision:


tX

Sn

3 238 3 25

11764

82. .

..t

XSn

3 238 3 25

11764

82. .

..


t0 2.947-2.947

.005


.005

t0 2.947-2.947

.005


.005

8 - 8 - 188188



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 16 - 1 = 1516 - 1 = 15



Decision:Decision:


tX

Sn

3 238 3 25

11764

82. .

..t

XSn

3 238 3 25

11764

82. .

..


There is no evidence There is no evidence average is not 3.25average is not 3.25t0 2.947-2.947

.005


.005

t0 2.947-2.947

.005


.005

8 - 8 - 189189


One-Tailed t Test One-Tailed t Test of Mean (Small Sample)of Mean (Small Sample)

8 - 8 - 190190


One-Tailed t TestOne-Tailed t TestExample Example

Is the average capacity of Is the average capacity of batteries batteries at least 140 at least 140 ampere-hours? A random ampere-hours? A random sample of sample of 2020 batteries had batteries had a mean of a mean of 138.47138.47 & a & a standard deviation of standard deviation of 2.662.66. . Assume a normal Assume a normal distribution. Test at the distribution. Test at the .05.05 level.level.

8 - 8 - 191191


One-Tailed t Test One-Tailed t Test SolutionSolution

HH00: :

HHaa: :

==

df =df =



Decision:Decision:


8 - 8 - 192192



HH00: : = 140 = 140

HHaa: : < 140 < 140

= =

df = df =



Decision:Decision:


8 - 8 - 193193



HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19



Decision:Decision:


8 - 8 - 194194


t0-1.729

.05

Reject

t0-1.729

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19



Decision:Decision:


8 - 8 - 195195



HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19



Decision:Decision:


tX

Sn

138 47 140

2 6620

2 57.

..t

XSn

138 47 140

2 6620

2 57.

..

t0-1.729

.05

Reject

t0-1.729

.05

Reject

8 - 8 - 196196



HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19



Decision:Decision:


tX

Sn

138 47 140

2 6620

2 57.

..t

XSn

138 47 140

2 6620

2 57.

..


t0-1.729

.05

Reject

t0-1.729

.05

Reject

8 - 8 - 197197



HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

df = df = 20 - 1 = 1920 - 1 = 19



Decision:Decision:


tX

Sn

138 47 140

2 6620

2 57.

..t

XSn

138 47 140

2 6620

2 57.

..


There is evidence pop. There is evidence pop. average is less than 140average is less than 140t0-1.729

.05

Reject

t0-1.729

.05

Reject

8 - 8 - 198198


One-Tailed t TestOne-Tailed t Test Thinking Challenge Thinking Challenge

You’re a marketing analyst for You’re a marketing analyst for Wal-Mart. Wal-Mart had teddy Wal-Mart. Wal-Mart had teddy bears on sale last week. The bears on sale last week. The weekly sales ($ 00) of bears weekly sales ($ 00) of bears sold in sold in 1010 stores was: stores was: 8 11 0 8 11 0 4 7 8 10 5 8 34 7 8 10 5 8 3. . At the At the .05.05 level, is there level, is there evidence that the average bear evidence that the average bear sales per store is sales per store is moremore thanthan 5 5 ($ 00)?($ 00)?


8 - 8 - 199199


One-Tailed t Test One-Tailed t Test Solution*Solution*

HH00: :

HHaa: :

= =

df = df =



Decision:Decision:


8 - 8 - 200200



HH00: : = 5 = 5

HHaa: : > 5 > 5

= =

df =df =



Decision:Decision:


8 - 8 - 201201



HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9



Decision:Decision:


8 - 8 - 202202


t0 1.833

.05

Reject

t0 1.833

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9



Decision:Decision:


8 - 8 - 203203



HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9



Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.

t0 1.833

.05

Reject

t0 1.833

.05

Reject

8 - 8 - 204204



HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9



Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.


t0 1.833

.05

Reject

t0 1.833

.05

Reject

8 - 8 - 205205



HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

df = df = 10 - 1 = 910 - 1 = 9



Decision:Decision:


tX

Sn

6 4 5

3 37310

131..

.tX

Sn

6 4 5

3 37310

131..

.


There is no evidence There is no evidence average is more than 5average is more than 5t0 1.833

.05

Reject

t0 1.833

.05

Reject

8 - 8 - 206206


Z Test of ProportionZ Test of Proportion

8 - 8 - 207207



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

LargeLargeSampleSample

Z Test(1 & 2tail)

Mean ProportionSmallSmallSampleSample

8 - 8 - 208208


One-Sample Z Test One-Sample Z Test for Proportionfor Proportion

8 - 8 - 209209



1.1. AssumptionsAssumptions Two categorical outcomesTwo categorical outcomes Population follows binomial distributionPopulation follows binomial distribution Normal approximation can be usedNormal approximation can be used

does not contain 0 or ndoes not contain 0 or nnp np p 3 1b g np np p 3 1b g

8 - 8 - 210210



1.1. AssumptionsAssumptions Two categorical outcomesTwo categorical outcomes Population follows binomial distributionPopulation follows binomial distribution Normal approximation can be usedNormal approximation can be used

does not contain 0 or ndoes not contain 0 or n

2.2. Z-test statistic for proportionZ-test statistic for proportion

Zp p

p pn

( )0

0 01Z

p pp p

n

( )0

0 01Hypothesized Hypothesized population proportionpopulation proportion

np np p 3 1b g np np p 3 1b g

8 - 8 - 211211


One-Proportion Z Test One-Proportion Z Test

Example Example The present packaging The present packaging system produces system produces 10%10% defective cereal boxes. defective cereal boxes. Using a new system, a Using a new system, a random sample of random sample of 200200 boxes hadboxes had1111 defects. defects. Does the new system Does the new system produce produce fewerfewer defects? defects? Test at the Test at the .05.05 level. level.

8 - 8 - 212212


One-Proportion Z One-Proportion Z Test SolutionTest Solution

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:


8 - 8 - 213213



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= =

nn = =



Decision:Decision:


8 - 8 - 214214



HH00: : pp = =.10.10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


8 - 8 - 215215



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject

8 - 8 - 216216



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject

Zp p

p pn

( )

.

. ( . ).0

0 01

11200

10

10 1 10200

212Zp p

p pn

( )

.

. ( . ).0

0 01

11200

10

10 1 10200

212

8 - 8 - 217217



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05


Zp p

p pn

( )

.

. ( . ).0

0 01

11200

10

10 1 10200

212Zp p

p pn

( )

.

. ( . ).0

0 01

11200

10

10 1 10200

212

8 - 8 - 218218



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05


There is evidence new There is evidence new system < 10% defectivesystem < 10% defective

Zp p

p pn

( )

.

. ( . ).0

0 01

11200

10

10 1 10200

212Zp p

p pn

( )

.

. ( . ).0

0 01

11200

10

10 1 10200

212

8 - 8 - 219219


One-Proportion Z One-Proportion Z Test Thinking Test Thinking

ChallengeChallengeYou’re an accounting You’re an accounting manager. A year-end audit manager. A year-end audit showed showed 4%4% of transactions of transactions had errors. You implement had errors. You implement new procedures. A random new procedures. A random sample of sample of 500500 transactions transactions had had 2525 errors. Has the errors. Has the proportionproportion of incorrect of incorrect transactions transactions changedchanged at at the the .05.05 levellevel? ?


8 - 8 - 220220


One-Proportion Z One-Proportion Z Test Solution*Test Solution*

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:


8 - 8 - 221221



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= =

nn = =



Decision:Decision:


8 - 8 - 222222



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500



Decision:Decision:


8 - 8 - 223223



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

8 - 8 - 224224



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

Zp p

p pn

( )

.

. ( . ).0

0 01

25500

04

04 1 04500

114Zp p

p pn

( )

.

. ( . ).0

0 01

25500

04

04 1 04500

114

8 - 8 - 225225



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025


Zp p

p pn

( )

.

. ( . ).0

0 01

25500

04

04 1 04500

114Zp p

p pn

( )

.

. ( . ).0

0 01

25500

04

04 1 04500

114

8 - 8 - 226226



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025


There is evidence There is evidence proportion is still 4% proportion is still 4%

Zp p

p pn

( )

.

. ( . ).0

0 01

25500

04

04 1 04500

114Zp p

p pn

( )

.

. ( . ).0

0 01

25500

04

04 1 04500

114

8 - 8 - 227227


ConclusionConclusion

1.1. Distinguished types of hypotheses Distinguished types of hypotheses

2.2. Described hypothesis testing processDescribed hypothesis testing process

3.3. Explained Explained pp-value concept-value concept

4.4. Solved hypothesis testing problems Solved hypothesis testing problems based on a single samplebased on a single sample

8 - 8 - 228228


This Class...This Class...

1.1. What was the most important thing you What was the most important thing you learned in class today?learned in class today?

2.2. What do you still have questions about?What do you still have questions about?

3.3. How can today’s class be improved?How can today’s class be improved?

Please take a moment to answer the following questions in writing:

End of Chapter

Any blank slides that follow are blank intentionally.

8 - 1 © 1998 Prentice-Hall, Inc. Chapter 8 Inferences Based on a Single Sample: Tests of...

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