8 - 1 © 2000 Prentice-Hall, Inc. Statistics for Business and Economics Inferences Based on a Single Sample: Tests of Hypothesis Chapter 8

8 - 8 - 11

© 2000 Prentice-Hall, Inc.© 2000 Prentice-Hall, Inc.

Statistics for Business Statistics for Business and Economicsand Economics

Inferences Based on a Single Sample: Inferences Based on a Single Sample: Tests of HypothesisTests of Hypothesis

Chapter 8Chapter 8

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 p-Value ConceptExplain p-Value Concept

4.4. Solve Hypothesis Testing Problems Solve Hypothesis Testing Problems Based on a Single SampleBased on a Single Sample

5.5. Explain Power of a TestExplain Power of a Test

8 - 8 - 33


Statistical MethodsStatistical Methods

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

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 - 1111


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 (Pronounced H-oh) (Pronounced H-oh)

5.5. Specified as HSpecified as H00: : Some Numeric Value Some Numeric Value Specified with = Sign Even if Specified with = Sign Even if , or , or Example, HExample, H00: : 3 3

8 - 8 - 1212


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 - 1313


Identifying Identifying HypothesesHypotheses

StepsSteps1.1. Example Problem: Test That the Example Problem: Test That the

Population Mean Is Not 3Population Mean Is Not 3

2.2. StepsSteps State the Question Statistically (State the Question Statistically ( 3) 3) State the Opposite Statistically (State the Opposite Statistically ( = 3) = 3)

Must Be Mutually Exclusive & ExhaustiveMust Be Mutually Exclusive & Exhaustive Select the Alternative Hypothesis (Select the Alternative Hypothesis ( 3) 3)

Has the Has the , , <<, or , or > > SignSign State the Null Hypothesis (State the Null Hypothesis ( = 3) = 3)

8 - 8 - 1414


State the question statistically: State the question statistically: = 12 = 12

State the opposite statistically: State the opposite statistically: 12 12

Select the alternative hypothesis: Select the alternative hypothesis: HHaa: :

1212

State the null hypothesis: State the null hypothesis: HH00: : = 12 = 12

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

What Are the What Are the Hypotheses?Hypotheses?

8 - 8 - 1515


State the question statistically: State the question statistically: 12 12

State the opposite statistically: State the opposite statistically: = 12 = 12

Select the alternative hypothesis: Select the alternative hypothesis: HHaa: :

1212

State the null hypothesis: State the null hypothesis: HH00: : = 12 = 12

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


8 - 8 - 1616




Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 20 20

State the null hypothesis: State the null hypothesis: HH00: : 20 20

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


8 - 8 - 1717




Select the alternative hypothesis: Select the alternative hypothesis: HHaa: : 25 25

State the null hypothesis: State the null hypothesis: HH00: : 25 25

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


8 - 8 - 1818


Basic IdeaBasic Idea

8 - 8 - 1919



Sample Mean = 50 Sample Mean = 50

HH00HH00

Sampling DistributionSampling Distribution

8 - 8 - 2020





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

8 - 8 - 2121






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

20202020HH00HH00

8 - 8 - 2222






... 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

8 - 8 - 2323


Level of SignificanceLevel of Significance

1.1. ProbabilityProbability

2.2. Defines Unlikely Values of Sample Defines Unlikely Values of Sample Statistic if Null Hypothesis Is TrueStatistic if Null Hypothesis Is True Called Rejection Region of Sampling Called Rejection Region of Sampling

DistributionDistribution

3.3. Designated Designated (alpha)(alpha) Typical Values Are .01, .05, .10Typical Values Are .01, .05, .10

4.4. Selected by Researcher at StartSelected by Researcher at Start

8 - 8 - 2424


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

8 - 8 - 2525



HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -

Level of ConfidenceLevel of Confidence

8 - 8 - 2626



HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -


Observed sample statisticObserved sample statistic

8 - 8 - 2727



HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion


1 - 1 -


8 - 8 - 2828


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

8 - 8 - 2929



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 - 3030



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 -


Observed sample statisticObserved sample statistic

8 - 8 - 3131



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 - 3232



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 - 3333


Decision Making RisksDecision Making Risks

8 - 8 - 3434


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)

8 - 8 - 3535


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

Innocent Correct ErrorDo NotReject

H0

1 - Type IIError

()


Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

8 - 8 - 3636


Jury Trial H0 Test



False

Innocent Correct Error AcceptH0

1 - Type IIError

()


Type IError ()

Power(1 - )

Jury Trial H0 Test



False

Innocent Correct Error AcceptH0

1 - Type IIError

()


Type IError ()

Power(1 - )

Decision ResultsDecision Results

HH00: Innocent: Innocent

8 - 8 - 3737


& & Have an Have an Inverse RelationshipInverse Relationship

You can’t reduce both errors simultaneously!

8 - 8 - 3838


Factors Affecting Factors Affecting

1.1. True Value of Population ParameterTrue Value of Population Parameter Increases When Difference With Hypothesized Increases When Difference With Hypothesized

Parameter DecreasesParameter Decreases

2.2. Significance Level, Significance Level, Increases When Increases When DecreasesDecreases

3.3. Population Standard Deviation, Population Standard Deviation, Increases When Increases When Increases Increases

4.4. Sample Size, Sample Size, nn Increases When Increases When nn Decreases Decreases

8 - 8 - 3939


Hypothesis Testing Hypothesis Testing StepsSteps

8 - 8 - 4040


HH00 Testing Steps Testing Steps

8 - 8 - 4141



State HState H00

State HState Haa

Choose Choose

Choose Choose nn

Choose testChoose test

8 - 8 - 4242



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 Haa

Choose Choose

Choose Choose nn

Choose testChoose test

8 - 8 - 4343


One Population One Population TestsTests

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

Mean Proportion Variance

2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

8 - 8 - 4444


Two-Tailed Z Test Two-Tailed Z Test of Mean (of Mean ( Known) Known)

8 - 8 - 4545



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

8 - 8 - 4646


Two-Tailed Z Test Two-Tailed Z Test for Mean (for Mean ( Known) Known)

1.1. AssumptionsAssumptions Population Is Normally DistributedPopulation Is Normally Distributed If Not Normal, Can Be Approximated by If Not Normal, Can Be Approximated by

Normal Distribution (Normal Distribution (nn 30) 30)

2.2. Alternative Hypothesis Has Alternative Hypothesis Has Sign Sign

8 - 8 - 4747


Two-Tailed Z Test Two-Tailed Z Test for Mean (for Mean ( Known) Known)



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 - 4848


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 2525 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 - 4949


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 - 5050



HH00: : = 368 = 368

HHaa: : 368 368

nn



Decision:Decision:


8 - 8 - 5151



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525



Decision:Decision:


8 - 8 - 5252



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525



Decision:Decision:


Z0 1.96-1.96

.025

Reject H 0 Reject H 0

.025

Z0 1.96-1.96

.025


.025

8 - 8 - 5353



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 3681525

150.

.ZX

n

372 5 3681525

150.

.

8 - 8 - 5454



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 3681525

150.

.ZX

n

372 5 3681525

150.

.

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

8 - 8 - 5555



HH00: : = 368 = 368

HHaa: : 368 368

.05.05

nn 2525



Decision:Decision:


Z0 1.96-1.96

.025


.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 3681525

150.

.ZX

n

372 5 3681525

150.

.


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

8 - 8 - 5656


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?

8 - 8 - 5757


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

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:


8 - 8 - 5858



HH00: : = 70 = 70

HHaa: : 70 70

= =

nn = =



Decision:Decision:


8 - 8 - 5959



HH00: : = 70 = 70

HHaa: : 70 70

= = .05.05

nn = = 3636



Decision:Decision:


8 - 8 - 6060



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 - 6161



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 7 703 536

51.

..Z

X

n

69 7 703 536

51.

..

8 - 8 - 6262



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 7 703 536

51.

..Z

X

n

69 7 703 536

51.

..


8 - 8 - 6363



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 7 703 536

51.

..Z

X

n

69 7 703 536

51.

..


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

8 - 8 - 6464


One-Tailed Z Test One-Tailed Z Test of Mean (of Mean ( Known) Known)

8 - 8 - 6565


One-Tailed Z Test One-Tailed Z Test for Mean (for Mean ( Known) Known)



2.2. Alternative Hypothesis Has < or > SignAlternative Hypothesis Has < or > Sign

8 - 8 - 6666


One-Tailed Z Test One-Tailed Z Test for Mean (for Mean ( Known) Known)



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 - 6767


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

8 - 8 - 6868


Z0

Reject H 0

Z0

Reject H 0


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

Must be Must be significantlysignificantly below below

8 - 8 - 6969


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 - 7070


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

8 - 8 - 7171


Z0

= 1

Z0

= 1


What Is Z given What Is Z given = .025? = .025?

= .025= .025

8 - 8 - 7272


Z0

= 1

Z0

= 1


.500 .500 -- .025.025

.475.475


= .025= .025

8 - 8 - 7373


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 - 7474


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 - 7575


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 2525 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 - 7676


One-Tailed Z Test One-Tailed Z Test SolutionSolution

HH00: :

HHaa: :

= =

n n = =



Decision:Decision:


8 - 8 - 7777



HH00: : = 368 = 368

HHaa: : > 368 > 368

= =

n n = =



Decision:Decision:


8 - 8 - 7878



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


8 - 8 - 7979



HH00: : = 368 = 368

HHaa: : > 368 > 368

= = .05.05

n n = = 2525



Decision:Decision:


Z0 1.645

.05

Reject

Z0 1.645

.05

Reject

8 - 8 - 8080



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 3681525

150.

.ZX

n

372 5 3681525

150.

.

8 - 8 - 8181



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 3681525

150.

.ZX

n

372 5 3681525

150.

.


8 - 8 - 8282



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 3681525

150.

.ZX

n

372 5 3681525

150.

.


No evidence average No evidence average is more than 368is more than 368

8 - 8 - 8383


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 - 8484


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

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:



Decision:Decision:


8 - 8 - 8585



HH00: : = 32 = 32

HHaa: : < 32 < 32

= =

nn = =



Decision:Decision:



Decision:Decision:


8 - 8 - 8686



HH00: : = 32 = 32

HHaa: : < 32 < 32

== .01 .01

nn = = 6060



Decision:Decision:



Decision:Decision:


8 - 8 - 8787



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 - 8888



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 7 323 860

2 65.

..Z

X

n

30 7 323 860

2 65.

..

8 - 8 - 8989



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 7 323 860

2 65.

..Z

X

n

30 7 323 860

2 65.

..

Reject at Reject at = .01 = .01

8 - 8 - 9090



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 7 323 860

2 65.

..Z

X

n

30 7 323 860

2 65.

..


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

8 - 8 - 9191


Observed Significance Observed Significance Levels: p-ValuesLevels: p-Values

8 - 8 - 9292


p-Valuep-Value

1.1. Probability of Obtaining a Test Statistic Probability of Obtaining a Test Statistic More Extreme (More Extreme (or or than Actual than Actual Sample Value Given HSample Value 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 p-Value If p-Value , Do Not Reject H, Do Not Reject H00

If p-Value < If p-Value < , Reject H, Reject H00

8 - 8 - 9393


Two-Tailed Z Test Two-Tailed Z Test p-Value Example p-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 2525 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 p-Value.p-Value. 368 gm.368 gm.

8 - 8 - 9494


Two-Tailed Z Test Two-Tailed Z Test p-Value Solutionp-Value Solution

8 - 8 - 9595



Z0 1.50-1.50 Z0 1.50-1.50

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

ZX

n

372 5 3681525

150.

.ZX

n

372 5 3681525

150.

.

8 - 8 - 9696



Z0 1.50-1.50 Z0 1.50-1.50


p-value is P(Z p-value is P(Z -1.50 or Z -1.50 or Z 1.50) 1.50)

8 - 8 - 9797



Z0 1.50-1.50

1/2 p-Value1/2 p-Value

Z0 1.50-1.50




8 - 8 - 9898



Z0 1.50-1.50


Z0 1.50-1.50



From Z table: From Z table: lookup 1.50lookup 1.50

.4332.4332


8 - 8 - 9999



Z0 1.50-1.50


Z0 1.50-1.50




.4332.4332

.5000.5000-- .4332.4332

.0668.0668


8 - 8 - 100100



Z0 1.50-1.50

1/2 p-Value.0668

1/2 p-Value.0668

Z0 1.50-1.50

1/2 p-Value.0668

1/2 p-Value.0668

p-value is P(Z p-value is P(Z -1.50 or Z -1.50 or Z 1.50) 1.50) = = .1336.1336

Z value of sample Z value of sample statisticstatistic


.4332.4332

.5000.5000-- .4332.4332

.0668.0668

8 - 8 - 101101



0 1.50-1.50 Z

RejectReject

0 1.50-1.50 Z

RejectReject

1/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .0668

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

8 - 8 - 102102



0 1.50-1.50 Z

RejectReject

0 1.50-1.50 Z

RejectReject

(p-Value = .1336) (p-Value = .1336) ( ( = .05). = .05). Do not reject.Do not reject.

1/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .06681/2 p-Value = .0668

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

Test statistic is in ‘Do not reject’ regionTest statistic is in ‘Do not reject’ region

8 - 8 - 103103


One-Tailed Z Test One-Tailed Z Test p-Value Example p-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 2525 boxes showedboxes showedX = 372.5X = 372.5. . The company has The company has specified specified to be to be 2525 grams. Find the p-Value.grams. Find the p-Value. 368 gm.368 gm.

8 - 8 - 104104


One-Tailed Z Test One-Tailed Z Test p-Value Solutionp-Value Solution

8 - 8 - 105105



Z0 1.50 Z0 1.50


ZX

n

372 5 3681525

150.

.ZX

n

372 5 3681525

150.

.

8 - 8 - 106106



Z0 1.50

p-Value

Z0 1.50

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

p-Value is P(Z p-Value is P(Z 1.50) 1.50)


8 - 8 - 107107



Z0 1.50

p-Value

Z0 1.50





.4332.4332

8 - 8 - 108108



Z0 1.50

p-Value

Z0 1.50





.4332.4332

.5000.5000-- .4332.4332

.0668.0668

8 - 8 - 109109



Z0 1.50

p-Value.0668

Z0 1.50

p-Value.0668



.4332.4332

Use Use alternative alternative hypothesis hypothesis to find to find directiondirection

.5000.5000-- .4332.4332

.0668.0668

p-Value is P(Z p-Value is P(Z 1.50) = .0668 1.50) = .0668

8 - 8 - 110110



0 1.50 Z

Reject

0 1.50 Z

Reject

p-Value = .0668p-Value = .0668

= .05= .05

8 - 8 - 111111



0 1.50 Z

Reject

0 1.50 Z

Reject

(p-Value = .0668) (p-Value = .0668) ( ( = .05). = .05). Do not reject.Do not reject.

p-Value = .0668p-Value = .0668

= .05= .05

Test statistic is in ‘Do not reject’ regionTest statistic is in ‘Do not reject’ region

8 - 8 - 112112


p-Value p-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 (p-Valuep-Value)?)?

8 - 8 - 113113


p-Value p-Value Solution*Solution*

Z0-2.65

p-Value.004

Z0-2.65

p-Value.004

Z value of Z value of sample statisticsample statistic


.4960.4960

Use Use alternative alternative hypothesis hypothesis to find to find directiondirection

.5000.5000-- .4960.4960

.0040.0040

p-Value is P(Z p-Value is P(Z -2.65) = .004. -2.65) = .004.p-Value < (p-Value < ( = .01). Reject H = .01). Reject H00..

8 - 8 - 114114


Two-Tailed t Test Two-Tailed t Test of Mean (of Mean ( Unknown) Unknown)

8 - 8 - 115115



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

8 - 8 - 116116


t Test for Mean t Test for Mean (( Unknown) Unknown)

1.1. AssumptionsAssumptions Population Is Normally DistributedPopulation Is Normally Distributed If Not Normal, Only Slightly Skewed & If Not Normal, Only Slightly Skewed &

Large Sample (Large Sample (nn 30) Taken 30) Taken

2.2. Parametric Test ProcedureParametric Test Procedure

8 - 8 - 117117


t Test for Mean t Test for Mean (( Unknown) Unknown)

1.1. AssumptionsAssumptions Population Is Normally DistributedPopulation Is Normally Distributed If Not Normal, Only Slightly Skewed & If Not Normal, Only Slightly Skewed &

Large Sample (Large Sample (nn 30) Taken 30) Taken

2.2. Parametric Test ProcedureParametric Test Procedure

3.3. t Test Statistict Test Statistic

tX

Sn

tX

Sn

8 - 8 - 118118


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

ValuesValues

8 - 8 - 119119


t0 t0


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

8 - 8 - 120120


t0 t0


ValuesValues

/2 = .05/2 = .05

/2 = .05/2 = .05

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

8 - 8 - 121121


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 - 122122


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 - 123123


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 - 124124


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 3636 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 - 125125


Two-Tailed t Test Two-Tailed t Test SolutionSolution

HH00: :

HHaa: :

= =

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


Decision:Decision:


8 - 8 - 126126



HH00: : = 368 = 368

HHaa: : 368 368

= =

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


Decision:Decision:


8 - 8 - 127127



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05

df = df = 36 - 1 = 3536 - 1 = 35Critical Value(s):Critical Value(s):


Decision:Decision:


8 - 8 - 128128



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

8 - 8 - 129129



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

tX

Sn

372 5 3681236

2 25.

.tX

Sn

372 5 3681236

2 25.

.

8 - 8 - 130130



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

tX

Sn

372 5 3681236

2 25.

.tX

Sn

372 5 3681236

2 25.

.


8 - 8 - 131131



HH00: : = 368 = 368

HHaa: : 368 368

= = .05.05



Decision:Decision:


t0 2.0301-2.0301

.025


.025

t0 2.0301-2.0301

.025


.025

tX

Sn

372 5 3681236

2 25.

.tX

Sn

372 5 3681236

2 25.

.


There is evidence pop. There is evidence pop. average is not 368average is not 368

8 - 8 - 132132


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 6464 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 - 133133


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

HH00: :

HHaa: :

df df



Decision:Decision:


8 - 8 - 134134



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

df df



Decision:Decision:


8 - 8 - 135135



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63



Decision:Decision:


8 - 8 - 136136



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63



Decision:Decision:


t0 2.6561-2.6561

.005


.005

t0 2.6561-2.6561

.005


.005

8 - 8 - 137137



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63



Decision:Decision:


t0 2.6561-2.6561

.005


.005

t0 2.6561-2.6561

.005


.005

tX

Sn

3 238 3 2511764

82. .

..t

XSn

3 238 3 2511764

82. .

..

8 - 8 - 138138



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63



Decision:Decision:


t0 2.6561-2.6561

.005


.005

t0 2.6561-2.6561

.005


.005

tX

Sn

3 238 3 2511764

82. .

..t

XSn

3 238 3 2511764

82. .

..


8 - 8 - 139139



HH00: : = 3.25 = 3.25

HHaa: : 3.25 3.25

.01.01

df df 64 - 1 = 6364 - 1 = 63



Decision:Decision:


t0 2.6561-2.6561

.005


.005

t0 2.6561-2.6561

.005


.005

tX

Sn

3 238 3 2511764

82. .

..t

XSn

3 238 3 2511764

82. .

..


There is no evidence There is no evidence average is not 3.25average is not 3.25

8 - 8 - 140140


One-Tailed t Test One-Tailed t Test of Mean (of Mean ( Unknown) Unknown)

8 - 8 - 141141


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 - 142142


One-Tailed t Test One-Tailed t Test SolutionSolution

HH00: :

HHaa: :

==

df =df =



Decision:Decision:


8 - 8 - 143143



HH00: : = 140 = 140

HHaa: : < 140 < 140

= =

df = df =



Decision:Decision:


8 - 8 - 144144



HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


8 - 8 - 145145


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


8 - 8 - 146146


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


tX

Sn

138 47 1402 66

20

2 57.

..t

XSn

138 47 1402 66

20

2 57.

..

8 - 8 - 147147


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


tX

Sn

138 47 1402 66

20

2 57.

..t

XSn

138 47 1402 66

20

2 57.

..


8 - 8 - 148148


t0-1.7291

.05

Reject

t0-1.7291

.05

Reject


HH00: : = 140 = 140

HHaa: : < 140 < 140

= = .05.05

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



Decision:Decision:


tX

Sn

138 47 1402 66

20

2 57.

..t

XSn

138 47 1402 66

20

2 57.

..


There is evidence pop. There is evidence pop. average is less than 140average is less than 140

8 - 8 - 149149


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 - 150150


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

HH00: :

HHaa: :

= =

df = df =



Decision:Decision:


8 - 8 - 151151



HH00: : = 5 = 5

HHaa: : > 5 > 5

= =

df =df =



Decision:Decision:


8 - 8 - 152152



HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


8 - 8 - 153153


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


8 - 8 - 154154


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


tX

Sn

6 4 53 373

10

131..

.tX

Sn

6 4 53 373

10

131..

.

8 - 8 - 155155


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


tX

Sn

6 4 53 373

10

131..

.tX

Sn

6 4 53 373

10

131..

.


8 - 8 - 156156


t0 1.8331

.05

Reject

t0 1.8331

.05

Reject


HH00: : = 5 = 5

HHaa: : > 5 > 5

= = .05.05

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



Decision:Decision:


tX

Sn

6 4 53 373

10

131..

.tX

Sn

6 4 53 373

10

131..

.


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

8 - 8 - 157157


Z Test of ProportionZ Test of Proportion

8 - 8 - 158158


Data TypesData Types

Data

Numerical Qualitative

Discrete Continuous

Data

Numerical Qualitative

Discrete Continuous

8 - 8 - 159159


Qualitative DataQualitative Data

1.1. Qualitative Random Variables Yield Qualitative Random Variables Yield Responses That ClassifyResponses That Classify e.g., Gender (Male, Female)e.g., Gender (Male, Female)

2.2. Measurement Reflects # in CategoryMeasurement Reflects # in Category

3.3. Nominal or Ordinal ScaleNominal or Ordinal Scale

4.4. ExamplesExamples Do You Own Savings Bonds? Do You Own Savings Bonds? Do You Live On-Campus or Off-Campus?Do You Live On-Campus or Off-Campus?

8 - 8 - 160160


ProportionsProportions

1.1. Involve Qualitative VariablesInvolve Qualitative Variables

2.2. Fraction or % of Population in a CategoryFraction or % of Population in a Category

3.3. If Two Qualitative Outcomes, Binomial If Two Qualitative Outcomes, Binomial DistributionDistribution Possess or Don’t Possess CharacteristicPossess or Don’t Possess Characteristic

8 - 8 - 161161


ProportionsProportions

1.1. Involve Qualitative VariablesInvolve Qualitative Variables

2.2. Fraction or % of Population in a CategoryFraction or % of Population in a Category

3.3. If Two Qualitative Outcomes, Binomial If Two Qualitative Outcomes, Binomial DistributionDistribution Possess or Don’t Possess CharacteristicPossess or Don’t Possess Characteristic

4.4. Sample Proportion (Sample Proportion (pp))

pxn

number of successes

sample sizep

xn

number of successes

sample size

^̂

8 - 8 - 162162


pp

1.1. Approximated by Approximated by Normal Normal

DistributionDistribution

Excludes 0 or nExcludes 0 or n

2.2. MeanMean

3.3. Standard ErrorStandard Error

Sampling Sampling Distribution Distribution of Proportionof Proportion

P p P p


where where pp00 = Population Proportion = Population Proportionpp̂̂pp

nn

11

.0.0

.1.1

.2.2

.3.3

.0.0 .2.2 .4.4 .6.6 .8.8 1.01.0

PP^̂

P(PP(P^̂ )) ˆ1ˆ3ˆ ppnpn ˆ1ˆ3ˆ ppnpn

00 00

8 - 8 - 163163


Z Z = 0= 0

zz= 1= 1

ZZ

Standardizing Standardizing Sampling Distribution Sampling Distribution

of Proportionof Proportion

Sampling Sampling DistributionDistribution

Standardized Standardized Normal DistributionNormal Distribution

PP^̂PP

PP

ZZpp pp pp

pp pp

nn

^̂pp

pp

^̂

^̂

(( ))11

^̂

^̂

^̂00

00 00

8 - 8 - 164164



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

8 - 8 - 165165


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

8 - 8 - 166166



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 ˆ1ˆ3ˆ ppnpn ˆ1ˆ3ˆ ppnpn

8 - 8 - 167167



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

ˆ1ˆ3ˆ ppnpn ˆ1ˆ3ˆ ppnpn

8 - 8 - 168168


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 - 169169


One-Proportion Z One-Proportion Z Test SolutionTest Solution

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:


8 - 8 - 170170



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= =

nn = =



Decision:Decision:


8 - 8 - 171171



HH00: : pp = =.10.10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


8 - 8 - 172172



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 - 173173



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 - 174174



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject Reject at Reject at = .05 = .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 - 175175



HH00: : pp = .10 = .10

HHaa: : pp < .10 < .10

= = .05.05

nn = = 200200



Decision:Decision:


Z0-1.645

.05

Reject

Z0-1.645

.05

Reject Reject at Reject at = .05 = .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 - 176176


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 - 177177


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

HH00: :

HHaa: :

= =

nn = =



Decision:Decision:


8 - 8 - 178178



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= =

nn = =



Decision:Decision:


8 - 8 - 179179



HH00: : pp = .04 = .04

HHaa: : pp .04 .04

= = .05.05

nn = = 500500



Decision:Decision:


8 - 8 - 180180



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 - 181181



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 - 182182



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 - 183183



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 - 184184


Chi-Square (Chi-Square (22) Test ) Test of Varianceof Variance

8 - 8 - 185185



OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

8 - 8 - 186186


Chi-Square (Chi-Square (22) Test) Testfor Variancefor Variance

1.1. Tests One Population Variance or Tests One Population Variance or Standard DeviationStandard Deviation

2.2. Assumes Population Is Approximately Assumes Population Is Approximately Normally DistributedNormally Distributed

3.3. Null Hypothesis Is HNull Hypothesis Is H00: : 22 = = 0022

8 - 8 - 187187


Chi-Square (Chi-Square (22) Test) Testfor Variancefor Variance

1.1. Tests One Population Variance or Tests One Population Variance or Standard DeviationStandard Deviation

2.2. Assumes Population Is Approximately Assumes Population Is Approximately Normally DistributedNormally Distributed

3.3. Null Hypothesis Is HNull Hypothesis Is H00: : 22 = = 0022

4.4. Test StatisticTest Statistic

Hypothesized Pop. VarianceHypothesized Pop. Variance

Sample VarianceSample Variance

22

22

22

1)1)

(n(n SS

00

8 - 8 - 188188


Chi-Square (Chi-Square (22) ) DistributionDistribution

Select simple randomsample, size n.

Compute s2

Compute 2 =(n-1)s 2/2

Astronomical numberof 2 values

PopulationSampling Distributionsfor Different SampleSizes

21 2 30

Select simple randomsample, size n.

Compute s2

Compute 2 =(n-1)s 2/2

Astronomical numberof 2 values

PopulationSampling Distributionsfor Different SampleSizes

21 2 30

8 - 8 - 189189


Finding Critical Finding Critical Value ExampleValue Example

What is the critical What is the critical 22 value given: value given:HHaa: : 22 > 0.7 > 0.7

nn = 3 = 3 =.05? =.05?

8 - 8 - 190190


Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 0.010 … 0.103 … 5.991



20 20

22 Table Table (Portion)(Portion)


nn = 3 = 3 =.05? =.05?

8 - 8 - 191191





20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

8 - 8 - 192192





20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

8 - 8 - 193193





20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

8 - 8 - 194194





20

Reject

20

Reject

= .05= .05



nn = 3 = 3 =.05? =.05?

8 - 8 - 195195





20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

8 - 8 - 196196





20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

8 - 8 - 197197





20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

8 - 8 - 198198





20

Reject

20

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

8 - 8 - 199199





20 5.991

Reject

20 5.991

Reject

= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2


nn = 3 = 3 =.05? =.05?

8 - 8 - 200200



What Do You Do If the Rejection Region Is on the Left?

What Do You Do If the Rejection Region Is on the Left?

What is the critical What is the critical 22 value given: value given:HHaa: : 22 << 0.7 0.7

nn = 3 = 3 =.05? =.05?

8 - 8 - 201201





20 20



nn = 3 = 3 =.05? =.05?

8 - 8 - 202202


20 20


nn = 3 = 3 =.05? =.05?




= .05= .05


RejectReject

8 - 8 - 203203


20 20


nn = 3 = 3 =.05? =.05?




= .05= .05


RejectReject Upper Tail Area Upper Tail Area for Lower Critical for Lower Critical Value = 1-.05 = .95Value = 1-.05 = .95

8 - 8 - 204204


20 20


nn = 3 = 3 =.05? =.05?




= .05= .05



8 - 8 - 205205


20 20


nn = 3 = 3 =.05? =.05?




= .05= .05



dfdf = = nn - 1 = 2 - 1 = 2

8 - 8 - 206206


20 20


nn = 3 = 3 =.05? =.05?




= .05= .05



dfdf = = nn - 1 = 2 - 1 = 2

8 - 8 - 207207


20 .103 20 .103


nn = 3 = 3 =.05? =.05?




= .05= .05


dfdf = = nn - 1 = 2 - 1 = 2

Upper Tail Area Upper Tail Area for Lower Critical for Lower Critical Value = 1-.05 = .95Value = 1-.05 = .95

RejectReject

8 - 8 - 208208


Chi-Square (Chi-Square (22) Test ) Test Example Example

Is the variation in boxes Is the variation in boxes of cereal, measured by of cereal, measured by the the variancevariance, equal to , equal to 1515 grams? A random grams? A random sample of sample of 2525 boxes had boxes had a standard deviation ofa standard deviation of 17.717.7 grams. Test at the grams. Test at the .05.05 level. level.

8 - 8 - 209209


2200

Chi-Square (Chi-Square (22) Test ) Test SolutionSolution

HH00: :

HHaa: :

= =

df = df =



Decision:Decision:


8 - 8 - 210210


2200


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= =

df = df =



Decision:Decision:


8 - 8 - 211211


2200


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


8 - 8 - 212212


2200


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


/2 = .025/2 = .025

8 - 8 - 213213


2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


/2 = .025/2 = .025

8 - 8 - 214214


2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


/2 = .025/2 = .025

2222

22

22

22

1)1) (25 -(25 -1)1) 1717 77

1515

3333 4242

(n(n SS

00

..

..

8 - 8 - 215215


2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


Do Not Reject at Do Not Reject at = .05 = .05 /2 = .025/2 = .025

2222

22

22

22

1)1) (25 -(25 -1)1) 1717 77

1515

3333 4242

(n(n SS

00

..

..

8 - 8 - 216216


2200 39.36439.36412.40112.401


HH00: : 22 = 15 = 15

HHaa: : 22 15 15

= = .05.05

df = df = 25 - 1 = 2425 - 1 = 24



Decision:Decision:


Do Not Reject at Do Not Reject at = .05 = .05

There Is No Evidence There Is No Evidence 22 Is Not 15 Is Not 15

/2 = .025/2 = .025

2222

22

22

22

1)1) (25 -(25 -1)1) 1717 77

1515

3333 4242

(n(n SS

00

..

..

8 - 8 - 217217


Calculating Type II Calculating Type II Error ProbabilitiesError Probabilities

8 - 8 - 218218


Power of TestPower of Test

1.1. Probability of Rejecting False HProbability of Rejecting False H00

Correct DecisionCorrect Decision

2.2. Designated 1 - Designated 1 -

3.3. Used in Determining Test AdequacyUsed in Determining Test Adequacy

4.4. Affected byAffected by True Value of Population ParameterTrue Value of Population Parameter Significance Level Significance Level Standard Deviation & Sample Size Standard Deviation & Sample Size nn

8 - 8 - 219219


XX00 = 368= 368

RejectRejectDo NotDo NotRejectReject

Finding PowerFinding PowerStep 1Step 1

Hypothesis:Hypothesis:HH00: : 00 368 368

HH11: : 00 < 368 < 368 = .05= .05

n =n =15/15/2525

DrawDraw

8 - 8 - 220220


XX11 = 360= 360

XX00 = 368= 368


Finding PowerFinding PowerSteps 2 & 3Steps 2 & 3


HH11: : 00 < 368 < 368

‘‘True’ Situation:True’ Situation: 11 = 360 = 360

= .05= .05

n =n =15/15/2525

DrawDraw

DrawDraw

SpecifySpecify

1-1-

8 - 8 - 221221


XX11 = 360= 360 363.065363.065

XX00 = 368= 368




HH11: : 00 < 368 < 368


065.363

25

1564.13680

n

ZX L

065.363

25

1564.13680

n

ZX L

= .05= .05

n =n =15/15/2525

DrawDraw

DrawDraw

SpecifySpecify

8 - 8 - 222222


XX11 = 360= 360 363.065363.065

XX00 = 368= 368




HH11: : 00 < 368 < 368


= .05= .05

n =n =15/15/2525

= .154= .154

1-1- =.846 =.846

DrawDraw

DrawDraw

SpecifySpecify

Z TableZ Table

065.363

25

1564.13680

n

ZX L

065.363

25

1564.13680

n

ZX L

8 - 8 - 223223


Power CurvesPower Curves

PowerPower PowerPower

PowerPower

Possible True Values for Possible True Values for 11 Possible True Values for Possible True Values for 11

Possible True Values for Possible True Values for 11

HH00: : 00 HH00: : 00

HH00: : = =00

= 368 in = 368 in

ExampleExample

8 - 8 - 224224


ConclusionConclusion

1.1. Distinguished Types of Hypotheses Distinguished Types of Hypotheses

2.2. Described Hypothesis Testing ProcessDescribed Hypothesis Testing Process

3.3. Explained p-Value ConceptExplained p-Value Concept

4.4. Solved Hypothesis Testing Problems Solved Hypothesis Testing Problems Based on a Single SampleBased on a Single Sample

5.5. Explained Power of a TestExplained Power of a Test

End of Chapter

Any blank slides that follow are blank intentionally.

Documents

8 - 1 © 2000 Prentice-Hall, Inc. Statistics for Business and Economics Inferences Based on a Single Sample: Tests of Hypothesis Chapter 8